Calculating the Stopping Distance of a Car

[SteliosVas]

Homework Statement

Hi everyone, this is the attempted problem.

A car starts from rest and accelerates at 5m/s^2 for 15 seconds. It than decelerates at -2m/s^2 for an unknown time (say t').

Calculate how long it takes for the car to stop

Homework Equations

I know I should be used, the constant equations of acceleration. E.g Integrate v = u + at but after this I get confused

The Attempt at a Solution

Initial velocity is 0, and velocity after 15 seconds is 75m/s.

Now after this where can I go?

 

[Nathanael]

Hello and welcome to physics forums.

If you are initially moving at 75 m/s with a constant acceleration of -2 m/s/s, how long will it take to stop?


(I, personally, like to imagine the velocity/time graph. The slope of this graph is the acceleration (-2). So if the graph starts out at v=75 (at t=0) with a slope of -2, when will it cross the time axis?)
(Sorry if that made it more confusing, I just like to imagine the graph of problems like this one.)

 

[SteliosVas]

Hi there thanks for the super quick reply.

So basically, it was a online quiz question, and it says the correct answer is 33.75 seconds, that is why I am not sure if there is a mistake there. Based on your above explanation and my calculations I obtained 37.5 seconds.



***Apologies to note ***

The acceleration of the car (based on the graph) is a sloping upwards linear line.

Does this mean something different?

 

[Satvik Pandey]

I think 37.5 seconds is correct.

 

[Nathanael]

The acceleration of the car (based on the graph) is a sloping upwards linear line.

Upwards? Doesn't a negative slope mean downwards?

So basically, it was a online quiz question, and it says the correct answer is 33.75 seconds, that is why I am not sure if there is a mistake there. Based on your above explanation and my calculations I obtained 37.5 seconds.

That is weird.. I agree, it should be 37.5 seconds

 

[SteliosVas]

Upwards? Doesn't a negative slope mean downwards?



That is weird.. I agree, it should be 37.5 seconds

I shall post a picture for clarity sake

 

[Nathanael]

I shall post a picture for clarity sake

The acceleration in the first 15 seconds was not a constant 5m/s (I thought it was from your original post)

The velocity after 15 seconds would not be 75 m/s, do you know what it would be?
(It's the area under the first 15 seconds of the graph)

 

[SteliosVas]

Well if I integrate a=5t in respect to time, I should get velocity as 2.5t^2. So making t=15, velocity = 562.5m/s

Now that I know that, where do I proceed from here?

 

[Nathanael]

Well if I integrate a=5t in respect to time, I should get velocity as 2.5t^2. So making t=15, velocity = 562.5m/s

Now that I know that, where do I proceed from here?

a does not equal 5t

a equals 5 when t=15 (and a equals zero when t=0, and a is linear), therefore $a=\frac{t}{3}$But who cares what a is all you have to do is use the formula for the area of a triangle; Area = 1/2 Base * Height

(You can integrate if you wish, you'll get the same answer)
Once you find the velocity after 15 seconds, tell me how long you think it will take to decelerate to zero

 

[SteliosVas]

The area is I get (1*2* 15 * 5 ) = 75m/s.
After that I have no idea.

So this is my understanding so far.

Acceleration at t=5, is 5m/s^2.

Sorry you said velocity at t=15 does not equal 15, would it equal 37.5ms/^2?

So given that velocity at t= 15 = 37.5m/s

And it decelerates at 2m/s^2 37.5/2= 18.75 + 15 initial seconds

Gives a grand total of 33.75 seconds.

Thank you so much!

 

[Nathanael]

Velocity at t=15 is 75m/s.

This is not true

The area is I get (1*2* 15 * 5 ) = 75m/s.

1*2*15*5=150

But I suspect you meant to say (1/2)*15*5 ? That is equal to 37.5

Spoiler

The velocity after 15 seconds is 37.5 m/s

Edit:

So given that velocity at t= 15 = 37.5m/s

And it decelerates at 2m/s^2 37.5/2= 18.75 + 15 initial seconds

Gives a grand total of 33.75 seconds.

Thank you so much!

Good