Constant acceleration of sports car

In summary, you hit the brakes to get a 7.5m/s² deceleration, but you still hit the dog. You use kinematics equations to calculate the distance and time it took for the car to stop.
  • #1
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Homework Statement




You are driving your new sports car at a velocity of 90km/hr , when you suddenly see a dog step into the road 50m ahead . You hit the breaks hard to get maximum deceleration of 7.5m/s² .



Homework Equations



i)How far will you go before stopping?

ii)Can you avoid hitting the dog?

can anyone help me ? please ?
 
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  • #2
What equations do you know that involve kinematics?
 
  • #3
the topic was constant acceleration .. then i hardly to understand .. it means i can't understand my teacher .. not only me .. >.<
 
  • #4
Well the kinematic equations can be found on this webpage. Then read the paragraph that follows the equations, and try to understand them well because you'll need to use them a lot. These equations ONLY apply when acceleration is constant.

For this question you are told the:
Initial velocity, vi = 90 km/h
Final velocity, vf = 0 km/h (because you've stopped)
Acceleration, a = -7.5 m/s^2

So look for an equation that has all three of these values in order to get something useful :). Be careful with the units, and make sure you are consistent with length and time (i.e. make sure you always use meters and seconds, or kilometers and hours).
 
  • #5
my answer was -150m .. and the true answer was 42m only ..

I think my brain will be explode ..

can you show how to solve EmittingLight ?

please ?
 
  • #6
It's better if you show your working so that we can show where you went wrong.

I suspect it might be unit conversions, which you can check with tools like Wolfram Alpha. The initial velocity is 90 km/h whereas the acceleration uses meters and seconds so you have to convert one of them to the other's units. I like using SI units, so I'd convert to meters and seconds. In Wolfram Alpha you would type something like this. With all the unit conversions you should get:

Initial velocity, vi = 25 m/s
Final velocity, vf = 0 m/s (because you've stopped)
Acceleration, a = -7.5 m/s^2

If you look at the equations there are only two that have all of these variables:
[tex]v_{f}^2=v_{i}^2+2*a*d[/tex][tex]v_{f}=v_{i}+a*t[/tex]You know how to solve for t and d in the equations right? t stands for the time elapsed and d stands for the distance traveled. Do you know which equation you want then?
 
  • #7
to solve for t:

i used the second eq.

for d:

i used the first eq .

my t is -17.5s

then to my d is -610m

then I divide the both to have velocity ..

what's next ? :?
 
  • #8
Three of the values are known, I've told you what they are in all my previous posts: [itex]v_f, v_i[/itex] and [itex]a[/itex].

I'll put these into the equations for you:[tex]0^2=25^2+2*-7.5*d[/tex][tex]0=25+-7.5*t[/tex]Do you know how to solve for t and d now? Remember, they are all in meters and seconds now, since that's what I've converted them into.

Would you be able to show your working? Like show how you solved for t and d, so I can tell you what you did incorrectly.

EDIT: Nevermind I know what you're doing. You're not doing the algebra correctly. Do you know that if:
[itex]a=b*x[/itex]
That:
[itex]\frac{a}{b}=x[/itex]?

So if:
[itex]-625=2*-7.5*d[/itex]
Then:
[itex]\frac{-625}{2*-7.5}=d[/itex]
 
Last edited:
  • #9
0^2=25^2+2∗−7.5∗d
0=625-15d
(transfer:)
15d=625
divide both sides by 15 to remain the d
then, d=41.67m

for the second:
0=25+-7.5*t
0=25-7.5t
(transfer:)
7.5t=25
divide both sides by 7.5 to remain t:
t=10/3s

is it wrong ?
 
  • #10
Well, didn't you say that the true answer was 42m before? It looks like it was rounded up from 41.67m to me.

Do you know why you got t=-17.5 and d=-610 before? And not now? Based on what I've deduced, you actually already put in the correct values before.

EDIT:
I just thought I should add that since you only wanted d, you didn't need to use the second equation. That's why I asked before if you 'knew which equation to use'; you wanted to find d, so you needed to use the first equation. But look on the bright side of this extra confusion, now you know how long it took the car to stop! And yes, you have the correct answers now as far as I know.
 
Last edited:
  • #11
I think I'm going to crazy .. but so very thank you .. now i know how the constant acceleration ..
 

What is constant acceleration?

Constant acceleration refers to the rate of change of an object's velocity that remains the same over time. In other words, the object's speed is changing by the same amount every second.

How does a sports car achieve constant acceleration?

A sports car can achieve constant acceleration by applying a constant force to the car. This force, known as the engine's torque, pushes the car forward and causes it to accelerate. The amount of acceleration depends on the magnitude of the force and the car's mass.

Is constant acceleration the same as constant speed?

No, constant acceleration and constant speed are not the same. Constant speed means that an object is moving at a constant rate without changing its velocity. Constant acceleration, on the other hand, means that the object's velocity is changing at a constant rate.

How does constant acceleration affect a sports car's performance?

Constant acceleration can greatly impact a sports car's performance. It allows the car to reach high speeds quickly and efficiently, making it ideal for racing. Additionally, constant acceleration helps the car maintain a steady and smooth movement, allowing for better control and handling.

What factors can affect a sports car's constant acceleration?

The main factors that can affect a sports car's constant acceleration are engine power, weight, and aerodynamics. A more powerful engine can provide a greater force for acceleration, while a lighter car will have less mass to accelerate. Aerodynamics, such as air resistance, can also play a role in how quickly a sports car can achieve constant acceleration.

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