# Describe how acceleration varies kinematics problem

• late347
In summary, the problem involves a bicyclist increasing their velocity from 18km/h to 23km/h in 1.2 seconds, with a combined mass of 78kg. The power needed for this acceleration is assumed to be constant and is calculated to be approximately 510 Watts. The boundaries for the acceleration are determined by examining the forces and velocities at different instances, and it is found that the force and acceleration vary between 102N and 79.8371N, and 1.3076 m/s^2 and 1.0235 m/s^2, respectively. The formula used for this calculation is P = F * v, where P is power, F is force, and v is velocity.
late347

## Homework Statement

bicyclist increases its velocity at constant power

a) how great is this power when the bicycle's velocity increases, from 18km/h to 23km/h, during the time of 1,2 seconds

b) Within which boundaries does the acceleration vary? The combined mass of bicyclist and his bicycle is 78kg

## Homework Equations

delta E= W

0,5 * m * v^2 = Ekin

## The Attempt at a Solution

a)

I assumed the same mass of bycyle as it was described in the later b) portion of the problem. Assume mass = 78kg

The problem looks like very difficult to solve, without assuming the mass to be 78kg. I wonder whether it is possible to calculate the power, without this beginning assumption.

transform velocities into m/s

v0= 5 m/s (nice that it divided evenly by 3,6!)

v1= 6,388 m/s

delta Ekinetic = Work

work = 616,4552 joules
assuming flat ground during travelling.

P=W/t

P= 616,4552 joules / 1,2 s

P= 513,71 Watts

roughly 510 Watts (correct answer according to book answer, in this sense)b) Within which boundaries does bicycles acceleration vary?

I have no idea how to strictly speaking answer the b) part

I suppose one could examine the velocities v0 and v1.

It is known that v0= 5m/s and v1= 6,388m/s

I backtracked in my textbook and found a formula which seems useful in this case. I must have glazed over this formula or simply misunderstood the formula earlier.

Ok according to my physics book one can calculate thusly

power at instant = (force at instant) * (velocity at instant)

We already calculated the power which was 510 Watts.

If we assume constant power, then we could use

P = F * v

510 W = F1* 5 m/s

510 W = F2 * 6,388 m/s

F1= 102 N
F2= 79, 8371 N

F= ma

Then we can know already the mass, and the forces F1 and F2.

We can calculate a1 and a2

F1= m*a1

F2= m * a2

a1= 102N / 78kg
= 1,3076 m/s^2

a2= 79,8371 N / 78kg
=1,0235 m/s^2

You could easily derive this formula,$$P = W / t$$
where W is work and t is time
Since (F and d are in the same direction) $$W = F d ~~\text{if the force is constant}$$
So $$P = \frac{F d}{t}$$
You know that ## \frac dt = v_{average}##
The formula becomes $$P = F v_{average}$$

if you want to find the power at specific point, you could take really small intervals of d and t. However a better way is to substitute ##V_{average} ## with the ##V## at that instance.

Now for this example it seems that the force is changing, But if you take a really small interval you could approximate that the force is constant and we get this small interval by substituting the instantaneous velocity into our equation.
So assuming the question means that the power is constant at every point your solution seems to be correct at least to me.

Last edited:
Biker said:
You could easily derive this formula,$$P = W / t$$
where W is work and t is time
Since (F and d are in the same direction) $$W = F d ~~\text{if the force is constant}$$
So $$P = \frac{F d}{t}$$
You know that ## \frac dt = v_{average}##
The formula becomes $$P = F v_{average}$$

if you want to find the power at specific point, you could take really small intervals of d and t. However a better way is to substitute ##V_{average} ## with the ##V## at that instance.

Now for this example it seems that the force is changing, But if you take a really small interval you could approximate that the force is constant and we get this small interval by substituting the instantaneous velocity into our equation.
So assuming the question means that the power is constant at every point your solution seems to be correct at least to me.

indeed the question stated that the power is constant throughout.

And the bicycle is accelerated with constant power from v0 to v1.

v0 =5m/s
v1= 6,388m/s

Would you say that, keeping in mind the pre-requisite... constant power throughout. It seems that the force would have to be changingso there is constant power throughout, and mass is also constant. Velocities are known, and the time to accelerate into that velocity is known.

If the power remains constant, then it means that every instance, has the same power I would say logically speaking.Especially when we lock the power at a constant value. And we lock the instant velocities which were given for us (v0, and v1). Evidently the bicyclists mass would also not change much during the physical exertion of cycling at least in these short instances.

## 1. How do you calculate acceleration in a kinematics problem?

In a kinematics problem, acceleration can be calculated by dividing the change in velocity by the change in time. This can be represented by the equation acceleration = (final velocity - initial velocity) / time.

## 2. What is the difference between average and instantaneous acceleration?

Average acceleration is the overall change in velocity over a certain time period, while instantaneous acceleration is the acceleration at a specific moment in time. Average acceleration can be calculated using the final and initial velocities, while instantaneous acceleration requires the use of calculus to find the derivative of the velocity function.

## 3. How does acceleration affect an object's motion in a kinematics problem?

Acceleration is a measure of how quickly an object's velocity is changing. In kinematics problems, acceleration can either cause an object to speed up, slow down, or change direction. This change in motion can be represented by the acceleration vector, which includes both the magnitude and direction of the acceleration.

## 4. What is the role of time in kinematics problems involving acceleration?

Time is a crucial factor in kinematics problems involving acceleration. Acceleration is a rate of change over time, so the amount of time an object experiences acceleration will determine the overall change in its velocity. Additionally, time can be used to calculate the displacement of an object using the equation displacement = (initial velocity * time) + (1/2 * acceleration * time^2).

## 5. Can acceleration be negative in a kinematics problem?

Yes, acceleration can be negative in a kinematics problem. Negative acceleration, also known as deceleration or retardation, occurs when an object's velocity decreases over time. This can happen when an object is slowing down in the direction of its motion or when an object is speeding up in the opposite direction of its motion.

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