# Calculating acceleration on different gears in vehicle

• johnpauldisec
In summary, acceleration is not simply determined by force over mass, but also takes into account mechanical advantage and the characteristics of the engine and vehicle such as wheel size, gearing, and torque. While Newton's theory states that acceleration is directly proportional to force and inversely proportional to mass, it is not always accurate as it does not consider the force over distance and the effects of different gears and rotational speeds. The power generated by the engine also plays a crucial role in acceleration. Calculating real acceleration requires taking into account various factors such as torque, engine speed, gearing, wheel diameter, and wheel velocity, and cannot simply be determined by multiplying force and distance or torque and rotational speed. In some cases, even with the same force and power,
johnpauldisec
Acceleration is not Force/mass, mechanical advantage (?)

I have a vehicle with wheels and electric motor that drives that wheels.
Those are characteristics of engine

Resultant torque and velocity at wheel will depend on radius of that wheel, as wheel work as gearing.
Let's say i will try with wheel A of 1m radius and wheel B of 2m radius.
On same engine speed vehicle with wheel A will have bigger torque&force at wheel but with lower wheel velocity, opposite will be on wheel B.
But the acceleration force will be same, as it depends on power from engine.

This guy called Newton said that acceleration is force/mass. But he is liar as its not possible.
Because in my example on wheel A force is two times bigger than in Wheel B, but both vehicles have same acceleration at same engine revs.

Also look here:

this picture is from https://en.wikipedia.org/wiki/Mechanical_advantage article

On Both bikes same force and power is applied on pedal,
the difference is gearing of the bikes. Bike on the left is making twice force/torque bike on right but on two times shorter distance which means two times smaller wheel speed.
Both bicycles encounter same acceleration. If i was using Newton's incorrect calculating of acceleration bike on left would have bigger acceleration. Also, i could have infinite acceleration with proper gearing...

How to calculate real acceleration? I can't just multiple force with distance or wheel velocity. What when the bike is stopped and i start to apply force on pedal? the wheel's speed and distance is 0 so would be the acceleration

I know now that wheel and gearing are machines that use Mechanical Advantage
And wheel size and gearing is just converting between more/less torque and more/less velocity of wheel (distance)
The power is the same, and actually acceleration is same too

but how to calculate acceleration if i know engine torque, revs, gearing, wheel diameter, wheel velocity
take in mind that the wheel velocity can be 0 (when vehicle is stopped)
also, engine can also be stopped as its electrical dc engine that is able to produce torque at 0 rpm

look at this too:

those are torques on wheel and rpm on wheels on different gears
we all know that its the power that actually accelerates vehicle
but stupid Newton tells me that its the force/torque
if he was right i could use such gearing that would make 1000000 torque on wheel from engine that generates 1 Watt and i would have unreal acceleration

as i told, can't just multiply torque with distance/velocity as vehicle and engine can be stopped

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torque is not force. Unit of force is Newton. Unit of torque is Newton-Meter. Torque is force times distance. So your leverage needs to be part of your acceleration calculation.

@meBigGuy
force times distance? the distance is wheel diameter,
WHAT if i use same wheel size but i attach differential/gear to my engine
speed torque engine_rpm wheel_rpm
vehicle A 10 100 200 50
vehicle B 20 100 400 100

torque is same but vehicle B is applying DOUBLE the power of A
and the acceleration is double
but forces are same

also
imagine same cars
they generate 300 torque but the second car is at double rotational speed than the first
the torque and force at wheel are same, but the second car is having double speed

if we ignore friction/air/drivetrain deceleration, what we get:
the second one is accelerating two times faster even that torque is same

what is gear doing? it can half the torque but double rotational speed
or opposite, double torque but half rotational speed

in each case power and resulting ACCELERATION is same
but forces on wheels is different
and the distance that force is applying on
and wheel velocity

but stupid Newton theory is not saying anything about it
it just says force/mass is acceleration which is ********

if that was true i would make a gearing that would generate 500000 torque and force from 1 Watt engine and it would accelerate as hell even if it weights 1000 kg

also, even on same gear
imagine an engine that generate same torque on all rotational speeds
so i could drive 2000 rpm with 1000 torque and then 4000 rpm with 1000 torque
FORCE ON WHEEL AND WHEELSPIN RISK would be SAME in both situations
but at 4000rpm power would be double and so the acceleration

but can't just multiple torque times rotational speed
because engine and vehicle can be stalled
and now i generate torque
power is 0
but the vehicle has to move

can't just multiple torque with gearing. because gearing is same when 1000 rpm and 2000 rpm on same gear, but on 2000rpm power is double

maybe i should do somethin like this
torque * (1.0+distance_in_metres_travelled_by_wheel) = acceleration
but i need scientific basics to calculate acceleration

also imagine this:
you are on bicycle with electrical motor
this motor generates constant 100 Watts of power
so the acceleration is same on all rpms, 100, 1000, 10000
but the torque is diminishing, SO IS THE CHANCE TO WHEELSPIN
you can easily wheelspin when starting or raise front wheel
but can't when big speed

also, HOW IS THE ACCELERATION SAME when force is diminishing with speed

Look at this picture

compare first and fourth pulley
the fourth have only 25N applied but its generating same force and acceleration as 100N in first pulley
I KNOW WORK and energy are same

but the fact is first and fourth pulley have different forces applied but same acceleration and power
but stupid Newton says that acceleration is DIRECTLY proportionally to force and inversely to mass
he haven't said a word about force over distance

i can't just multiple force with distance
because what if there is no distance at all?
for example imagine this pulley, let's say this weight on bottom is heavy and it stays on floor
now i apply force but this force is too small to overcome gravity
so there is no movement
BUT THE FORCE IS THERE

johnpauldisec said:
but stupid Newton says that acceleration is DIRECTLY proportionally to force and inversely to mass
he haven't said a word about force over distance
johnpauldisec said:
This guy called Newton said that acceleration is force/mass. But he is liar as its not possible.
I'm dying of laughter right now.

Force and torque are not the same thing.

Mandelbroth said:
I'm dying of laughter right now.

Force and torque are not the same thing.

The force on wheels is torque on wheel divided by wheel radius

Don't even try to suggest that car on higher gears generate same force on wheels (assuming same engine rpm and power is present). The higher gear the less force -> its very easy to wheel spin at first gear if you press pedal to floor but its very hard to wheel spin at 3rd gear when pedal pressed to floor
and if we ignore deceleration forces like air/rolling/drivetrain losses
that car would accelerate SAME FAST on 1st and 5th gear
but on 1st it is easy to wheelspin and on 5th very unlikely

ALSO, car when driving at 1000 rpm of engine with torque 100 on wheels will accelerate two times slower than same car at 2000 rpm of engine (and double rpm of wheels) with same torque (100)
(the power will be double)

but don't even try to tell me that force on wheel is torque on wheel mutiplied by wheel speed
why? because at 0 rpm (car & engine stopped) car would never be able to move as force would always be 0
and electric engines are capable of generating torque at 0 rpm

the fact is first and fourth pulley have different forces applied but same acceleration and power but stupid Newton says that acceleration is DIRECTLY proportionally to force and inversely to mass

Newton actually says...

http://en.wikipedia.org/wiki/Newton's_laws_of_motion

Second law: The acceleration of a body is directly proportional to, and in the same direction as, the net force acting on the body, and inversely proportional to its mass..

The force you refer to is not "acting on the body" being accelerated. In each case the force acting on the body is the same = 100N.

With an electric motor, torque decreases linearly as rpm increases. For the graph, torque = 3.2 (1 - rpm/600). As shown in the graph, power = torque x rpm x 2 x pi / 60. Power also equals force x speed, so force = power / speed = (torque x rpm x 2 x pi / 60) / v, where v is velocity. Assume bike's velocity is 2 m/s, then driven wheel force at 2 m/s versus rpm looks like:

Code:
  rpm  torque   power   force

0   3.200   0.000   0.000
50   2.933  15.359   7.679
100   2.667  27.925  13.963
150   2.400  37.699  18.850
200   2.133  44.680  22.340
250   1.867  48.869  24.435
300   1.600  50.265  25.133
350   1.333  48.869  24.435
400   1.067  44.680  22.340
450   0.800  37.699  18.850
500   0.533  27.925  13.963
550   0.267  15.359   7.679
600   0.000   0.000   0.000

Note that the maximum force occurs at the rpm corresponding to peak power. To optimize the gearing, you'd want the center point between shifts to be near the power peak. With a continously variable transmission, you'd want the motor rpm to be kept constant at 300 rpm regardless of the bikes speed.

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rcgldr said:
Note that the maximum force occurs at the rpm corresponding to peak power. To optimize the gearing, you'd want the center point between shifts to be near the power peak. With a continously variable transmission, you'd want the motor rpm to be kept constant at 300 rpm regardless of the bikes speed.

I agree that best power and ACCELERATION will be at 300 rpm.
But you can't equal force with power.
In your posted forces, at 0 rpm force is 0. So the car/bike would NEVER START MOVING. Which is not true.

Also, look:
imagine a sports car that has peak 400HP at 6000 rpm.
I put first gear, accelerate slowly until i reach 5700 rpm at engine. Now i press acceleration pedal to the floor, what will happen? Rear (driven) wheels will start to spin.
Now i put in 6th gear and accelerate so i have 5700 rpm. Which could be like 280km/h vehicle speed. At this moment i put acceleration pedal to the floor (actually for sure it was already at floor or near). What happens? NO WHEELSPIN, just acceleration.

Why is that? because 5th gear provide times lower torque that in first gear. And force on wheel is: torque on wheel * wheel diameter.
So the higher gear the force has to be lower.
Look at this picture i posted earlier:

those are torques and rotations of engine and wheel at each gear.

Now look back at bicycle:

The "engine" power is same in both bikes. Because of gearing, the bike on right is traveling 2 times faster that bike on left. But you see that force on wheel is different, and if you put both bikes on slippery surface, left one would wheelspin/crash when right one could still drive safely.
We can see that bike on right is "applying" force over 2 times distance left bike, whatever "applying force" means. Newton say that force is a force, dosen't matter distance/speed its making.

Wheel with axle is kind of machine with mechanical advantage https://en.wikipedia.org/wiki/Wheel_and_axle
so gearing convert between more/less torque and more/less speed
but what with acceleration? how to calculate it?
Can't multiple torque * rotational speed because at 0rpm bike would never move. Also why does wheelspin not occur on sportscar in 6th gear at full power and occurs at 1st gear? the force has to be lower at 6th gear than 1, but acceleration force is same.

johnpauldisec said:
But you can't equal force with power.
power = force x speed.
force = power / speed.

johnpauldisec said:
In your posted forces, at 0 rpm force is 0.
Note the assumption is that the bike is moving a 2 meters / second. This would require an infinitely tall gear to convert 0 rpm into 2 meters / second speed, so the driven wheel force would be zero.

Assume the driven wheel has a radius of 1 meter, then here is the previous table showing the gear ratio corresponding to 2 meters / second.

wheel rpm = motor rpm / gear
wheel torque = motor torque * gear
wheel force = wheel torque / (1 meter)

Code:
  rpm  torque   power   force    gear

0   3.200   0.000   0.000   0.000
50   2.933  15.359   7.679   2.618
100   2.667  27.925  13.963   5.236
150   2.400  37.699  18.850   7.854
200   2.133  44.680  22.340  10.472
250   1.867  48.869  24.435  13.090
300   1.600  50.265  25.133  15.708
350   1.333  48.869  24.435  18.326
400   1.067  44.680  22.340  20.944
450   0.800  37.699  18.850  23.562
500   0.533  27.925  13.963  26.180
550   0.267  15.359   7.679  28.798
600   0.000   0.000   0.000  31.416

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rcgldr said:
Note the assumption is that the bike is moving a 2 meters / second. This would require an infinitely tall gear to convert 0 rpm into 2 meters / second speed, so the driven wheel force would be zero.
But i don't want to have 0 rpm of engine at bike 2 m/s speed. I want to have my bike stopped and engine stopped. Now i apply electrical power to engine so it starts generating torque at 0rpm. What force will be applied on wheels/on bike? Will it start to move? If i use your calculations the bike will never start.

rcgldr said:
wheel torque = motor torque * gear
wheel force = wheel torque / (1 meter)
I understand that. And that means that the smaller gear will generate smaller torque and FORCE at wheels.
now imagine i have engine attached to my car. and that engine is generating constant 50HP from 0 to 20000 rpm (that means torque is diminishing with revs).
And i have gearbox taken from passanger car. I achieved my top speed of 100 km/h at full power on 3rd gear. NOW I CHANGE GEAR to 5 and apply full power. the power of engine is same, but the toque on wheel is smaller so is the force (!). does that mean my car will start to decelerate to ~60-70km/h ?! its not possible. power is a power. energy can't be lost.

I don't understand what is the difference when i apply force 100N over 10 metres vs over 1 metre vs without movement. A force is a force. And acceleration is F/m.

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johnpauldisec said:
And i have gearbox taken from passanger car. I achieved my top speed of 100 km/h at full power on 3rd gear. NOW I CHANGE GEAR to 5 and apply full power. the power of engine is same, but the toque on wheel is smaller so is the force (!). does that mean my car will start to decelerate to ~60-70km/h ?! its not possible. power is a power. energy can't be lost.

If you really achieve a top speed of only 100 km/h at full power in 3rd gear, then you will indeed find that if you switch to larger gear, that the car will slow down. (unless the engine was really far to the right on the downslope of the torque/rev curve, and the engine can produce much more torque in a higher gear).

There will be no power disappearing, but the engine simply can't produce full power at the lower revs. There's only so much gasoline that can go in an explosion, and there's only so much explosions. (2 per revolution for a 4 cylinder car), and if there aren't enough explosions, the engine can't produce full power.

EDIT: this goes for electric motors too, and even for human cyclists. There's a maximum torque that can be produced, so at low revs, you can't get the maximum power of the engine

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@willem2
if you read my full post:

johnpauldisec said:
now imagine i have engine attached to my car. and that engine is generating constant 50HP from 0 to 20000 rpm (that means torque is diminishing with revs).

this engine's power is constant 50HP no matter revolutions

johnpauldisec said:
@willem2
if you read my full post:

this engine's power is constant 50HP no matter revolutions

What you described in your last post was an engine that had to slow down in fifth, because it would produce less force in fifth than in 3rd gear. That's not compatible with constant power, no matter what the revolutions are. Such an engine is impossible anyway, because there's always a maximum torque, and therefore a minimum number of revolutions needed to get full power.

willem2 said:
What you described in your last post was an engine that had to slow down in fifth, because it would produce less force in fifth than in 3rd gear. That's not compatible with constant power, no matter what the revolutions are. Such an engine is impossible anyway, because there's always a maximum torque, and therefore a minimum number of revolutions needed to get full power.

The engine is electrical and its maximum torque is at 0 rpm, on bigger rpm it has lower torque so power remains constant. Full power is on any rpm on that engine.

And seems we misunderstood. In my car i was driving full power at 3rd gear and switched to a fifth gear. Vehicle speed was same, speed of engine dropped but power remained the same.
Then i asked a question, if on 5th gear power is same but force on wheel is lower, will the car slow down or keep the speed, if keep, why?

johnpauldisec said:
The engine is electrical and its maximum torque is at 0 rpm, on bigger rpm it has lower torque so power remains constant. Full power is on any rpm on that engine.

And seems we misunderstood. In my car i was driving full power at 3rd gear and switched to a fifth gear. Vehicle speed was same, speed of engine dropped but power remained the same.
Then i asked a question, if on 5th gear power is same but force on wheel is lower, will the car slow down or keep the speed, if keep, why?

If the power is the same, and the speed is the same, the wheel force is the same. Power and wheel torque are not independent of each other.

(That having been said, power is not constant for an electric motor - it hits its peak at about half the motor's maximum RPM, and falls off on either side of that value, which can be seen in your very first graph in this thread)

I really find it difficult to help people who act like they know more than centuries of science as opposed to people who want to try to understand where their understanding is flawed. If they are not looking for the flaws in their understanding they will continue to make the same mistaken assumptions over and over.

I may be wrong, but I don't think you understand the relationship between torque and force, and how gearing changes the effective distances in converting torque to force. There also seems to be some misunderstanding of the nature/limitations of the power supplied through pedaling.

Again, I may be misunderstanding your misunderstanding.

Take a tiny wheel powered by a chain supplying torque. The torque is acting through a small distance, so the force is large (torque = force X distance, so force = torque/distance). That means rapid acceleration. If the wheel is large, the force is less for the same torque, so there is smaller acceleration. With the small wheel, if you could maintain the torque at all rotational speeds you would continue to accellerate at the same high rate forever.

cjl said:
If the power is the same, and the speed is the same, the wheel force is the same. Power and wheel torque are not independent of each other.

Really? then what if my car and engine is stopped. And now i apply electric power to motor. This motor can generate torque at 0 rpm. So there is also force on wheel. But the power is 0 because rpm is 0.

johnpauldisec said:
The engine is electrical and its maximum torque is at 0 rpm, on bigger rpm it has lower torque so power remains constant.
Look at the graph in the first post. The torque decreases linearly, and the power is at a maximum at 300 rpm, so the power versus rpm is not constant.

rcgldr said:
Look at the graph in the first post. The torque decreases linearly, and the power is at a maximum at 300 rpm, so the power versus rpm is not constant.

Ok, but then how the car will start moving when its stalled and engine is stalled? if power is 0 at 0 rpm.

johnpauldisec said:
Ok, but then how the car will start moving when its stalled and engine is stalled? if power is 0 at 0 rpm.

That's why there has to be a clutch somewhere in the drivetrain. If you have an automatic transmission you can hold the car with the brakes while the motor is turning at low revs, generating torques that is opposed by the torque from the brakes. If you have a manual transmission, when the clutch is in the motor can turn, generating non-zero torque, while the car isn't moving.

Nugatory said:
That's why there has to be a clutch somewhere in the drivetrain. If you have an automatic transmission you can hold the car with the brakes while the motor is turning at low revs, generating torques that is opposed by the torque from the brakes. If you have a manual transmission, when the clutch is in the motor can turn, generating non-zero torque, while the car isn't moving.

Yes, but clutch is for combustion engines. When they stall they don't generate any torque and they need electric starters to accelerate revs until it rotates on its own.
If you have electrical engine you don't need a clutch, even had a radio controlled toy?
Or have you ride bicycle? when its stopped and you press pedal it starts moving, even that on start it has 0 power.

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johnpauldisec said:
Yes, but clutch is for combustion engines. When they stall they don't generate any torque and they need electric starters to accelerate revs until it rotates on its own.
If you have electrical engine you don't need a clutch, even had a radio controlled toy?
Or have you ride bicycle? when its stopped and you press pedal it starts moving, even that on start it has 0 power.

Ah - I think I see what you're asking: how can something that is at rest start moving if the power delivered while it's at rest is zero no matter how great the torque? The heuristic answer is that changes in speed are produced by forces (straight-line motion) and torques (rotary motion) so if there is net torque or force, there will be a change in speed; if the speed was zero before the force/torque was applied, it won't be zero afterwards. Power only comes into it when you start to calculate the distance covered by the force in a given amount of time.

Nugatory said:
Ah - I think I see what you're asking: how can something that is at rest start moving if the power delivered while it's at rest is zero no matter how great the torque? The heuristic answer is that changes in speed are produced by forces (straight-line motion) and torques (rotary motion) so if there is net torque or force, there will be a change in speed; if the speed was zero before the force/torque was applied, it won't be zero afterwards.
But its not so simple. Take in mind i can change torque to anything i want using gears. I could apply a gearing that would make from 100 Nm of torque at engine a 50000 Nm of torque at wheels. That can't mean i could accelerate fast as hell, there is no free energy

Nugatory said:
Power only comes into it when you start to calculate the distance covered by the force in a given amount of time.
But what is "distance covered by the force",
and what is difference when force of 100N apply on 70kg milf over distance of one metre in one second than when force of 100N apply on 70kg milf over distance of two metres in one second ?
if first milf had velocity of 1 m/s and second milf had velocity of 2 m/s, what will be resultant velocities after those forces applied in one second?
Or if its easier for you we can assume both milf's had 1 m/s of velocity (just write what speed did you choose)

Power = force x speed. If speed is zero, then force can be non-zero even though power is zero. This only lasts for an infinitesimal moment. No clutch is needed for electric motor, and no clutch is used in diesel electric locomotive, the engine drives a generator that in turn drives electric motors connected to the driven wheels.

johnpauldisec said:
But its not so simple. Take in mind i can change torque to anything i want using gears. I could apply a gearing that would make from 100 Nm of torque at engine a 50000 Nm of torque at wheels. That can't mean i could accelerate fast as hell, there is no free energy

You can get a very high acceleration that way, limited only because if you apply enough torque, something will break. However, a very high acceleration doesn't mean a very high final speed, not unless you can keep applying the torque for long enough to build up the speed.

Suppose you have an engine that delivers 100 Nm of torque at the crankshaft when operating at 6000 RPM at full throttle. You can gear it down to deliver 50000 Nm at the wheels, but then the wheels will only be turning at 12 RPM; for reasonable-sized wheels with a circumference of two meters that's about 1.4 km/hr. That's a bulldozer, something that can pull very hard but doesn't cover a lot of ground.

If it's light enough, its zero-to-1.4 acceleration will be pretty impressive. But if you want to use that impressive acceleration to win a drag race you'll need to either find a way of getting the motor to deliver the same torque at much higher rev rates (there's a reason why Formula 1 engines do more than 15000 RPM, and aircraft turbines spin even faster) or get the motor to deliver more torque at 6000 RPM so that you can get the same impressive acceleration with less aggressive gearing. Either way, you're asking the motor to generate more power (the product of torque and revs per unit time).

johnpauldisec said:
But what is "distance covered by the force",
and what is difference when force of 100N apply on 70kg milf over distance of one metre in one second than when force of 100N apply on 70kg milf over distance of two metres in one second ?

You're trying to specify the distance, the time, the force, and the mass all at once, and you can't do that because they're connected by Newton's laws. For example, the force and the mass determine the acceleration and the acceleration and the initial speed determine the time it takes to cover one meter; so once you've chosen a mass and a force, you're stuck with a particular time.

If you don't mind, I'm going to do this with a mass of 50 kg instead of 70, because I'm doing the arithmetic in my head.

Say the 50 kg mass starts at rest, and we apply 100N to it for a distance of one meter. The total work done is 100 Joules; this comes from ##W=Fd##. All this work ends up as kinetic energy of the mass and I can use the kinetic energy formula ##E=\frac{mv^2}{2} \Rightarrow v=\sqrt{2E/m}## to calculate that the mass is moving at a speed of ##2 m/sec## after it's covered the first meter. Its acceleration (from ##F=ma##) is ##2 m/sec^2## so I can calculate (from ##s=\frac{at^2}{2}##) that it took one second to cover that one meter distance. The power output was 100 Joules over a period of one second, which is 100 Watts.

There are two important points here:
1) That 100 watts is an average over the entire second. At the start of the second when the speed was zero, the instaneous power delivered was zero, and by the end of the second it was 200 watts. So if my motor wasn't capable of delivering 200 watts, it wouldn't have been able to keep applying the full 100N of force for the entire second.
2) if I'm going to continue applying 100N to the mass for another second, thereby maintaining the same ##2 m/sec^2## acceleration so that it is moving at ##4 m/sec##, I'll need an even more powerful motor. Don't take my word for it, calculate it... The work done in the second meter is the same (##W=Fd##) but the time it takes the mass to cover that meter is less because it's moving faster.

I made up some gear ratios for the motor shown in the first post to create a pair of graphs. The first graph shows driven wheel torque (or force if radius is 1 meter) versus speed for each gear, and the second graph shows driven wheel power versus speed for each gear. The ideal shift points occur where the lines cross (the lines cross at the same speeds in both graphs).

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## 1. How do you calculate acceleration on different gears in a vehicle?

To calculate acceleration on different gears in a vehicle, you need to know the vehicle's mass, the force produced by the engine, and the gear ratio. The formula for acceleration is acceleration = force / mass. The gear ratio can be found by dividing the number of teeth on the driven gear by the number of teeth on the driving gear.

## 2. How does gear ratio affect acceleration in a vehicle?

The gear ratio affects acceleration by determining how much force is applied to the wheels. A higher gear ratio means more force is applied, resulting in faster acceleration. Conversely, a lower gear ratio means less force is applied, resulting in slower acceleration.

## 3. How does the type of transmission impact acceleration?

The type of transmission can impact acceleration by changing the gear ratio. Manual transmissions allow the driver to manually shift gears, while automatic transmissions shift gears automatically. Manual transmissions typically have more gear options, allowing for better acceleration control, while automatic transmissions may have a more limited gear ratio range.

## 4. Are there any other factors that can affect acceleration on different gears?

Yes, there are other factors that can affect acceleration on different gears, such as the vehicle's weight distribution, tire traction, and road conditions. These factors can impact the amount of force that is applied to the wheels and therefore, the acceleration.

## 5. Can you calculate acceleration on different gears without knowing the vehicle's mass?

No, you cannot accurately calculate acceleration on different gears without knowing the vehicle's mass. Mass is an essential component of the acceleration formula and without it, the calculation will not be accurate. Therefore, it is important to know the vehicle's mass when calculating acceleration on different gears.

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