Calculating the Stopping Distance of a Car

In summary, the problem involves a car starting from rest and accelerating at 5m/s^2 for 15 seconds, and then decelerating at -2m/s^2 until it stops. The correct answer is 33.75 seconds. However, there was some confusion about the initial velocity and the acceleration, but it was eventually resolved using the formula for the area of a triangle. The final solution was found by adding the time taken for acceleration and the time taken for deceleration.
  • #1
SteliosVas
70
0

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?
 
Physics news on Phys.org
  • #2
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.)
 
  • #3
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?
 
  • #4
I think 37.5 seconds is correct.
 
  • #5
SteliosVas said:
The acceleration of the car (based on the graph) is a sloping upwards linear line.

Upwards? Doesn't a negative slope mean downwards?

SteliosVas said:
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
 
  • #6
Nathanael said:
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
 

Attachments

  • Screen Shot 2014-08-25 at 1.21.29 pm.png
    Screen Shot 2014-08-25 at 1.21.29 pm.png
    16.3 KB · Views: 497
  • #7
SteliosVas said:
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)
 
  • #8
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?
 
  • #9
SteliosVas said:
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 [itex]a=\frac{t}{3}[/itex]But who cares what a is :tongue: 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
 
  • #10
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!
 
Last edited:
  • #11
SteliosVas said:
Velocity at t=15 is 75m/s.
This is not true

SteliosVas said:
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

The velocity after 15 seconds is 37.5 m/s



Edit:
SteliosVas said:
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 :smile:
 
Last edited:
  • Like
Likes 1 person

1. What is stopping distance?

Stopping distance is the distance a car travels from the moment the brakes are applied until the car comes to a complete stop.

2. What factors affect stopping distance?

Stopping distance is affected by a variety of factors, including the speed of the car, the condition of the road surface, the condition of the tires, and the driver's reaction time.

3. How is stopping distance calculated?

Stopping distance can be calculated by adding the reaction distance (the distance the car travels while the driver reacts to a hazard) and the braking distance (the distance the car travels while braking until it comes to a stop).

4. How does speed impact stopping distance?

The higher the speed of the car, the longer the stopping distance will be. This is because the car will have more momentum and require more time and distance to come to a complete stop.

5. How can stopping distance be reduced?

Stopping distance can be reduced by maintaining a safe following distance, keeping tires properly inflated, and regularly checking and replacing brake pads. Additionally, driving at a safe and appropriate speed can also help reduce stopping distance.

Similar threads

  • Introductory Physics Homework Help
Replies
13
Views
669
  • Introductory Physics Homework Help
Replies
6
Views
589
  • Introductory Physics Homework Help
2
Replies
57
Views
537
  • Introductory Physics Homework Help
Replies
7
Views
1K
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
454
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
942
  • Introductory Physics Homework Help
Replies
1
Views
7K
Back
Top