# Homework Help: Find total displacement of an automobile

1. Jun 18, 2014

1. The problem statement, all variables and given/known data
An automobile travels in a straight line for 10 seconds at 20 m/s than accelerates uniformly to a speed of 30 m/s in the next 10 seconds. Find the total displacement.

2. Relevant equations

d = 1/2(v0+v)t

3. The attempt at a solution

v0 = 20 m/s

v = 30 m/s

t = t1 + t2 = 10 sec + 10 sec = 20 sec

d = 1/2 (20 + 30) (20)

d = 500 m

The answer key though says it should be 450 m and I'm not sure why that would be the case...

2. Jun 18, 2014

### ChiralWaltz

Is the car at a constant velocity while it is accelerating?

3. Jun 18, 2014

No, its velocity changes when it accelerates, right? It's increasing its velocity by 10 m/s I think..

4. Jun 18, 2014

I don't think I fully understand this problem...

5. Jun 18, 2014

### ChiralWaltz

Yes. Acceleration is the change in velocity. In the first 10 seconds, what is constant? In the second ten second, what is constant?

Edit: that was incredibly vague. If the car travels 20 m/s for 10 seconds, how far does it go?

The car is now traveling 20 m/s. It accelerates (this is your a) from 20 m/s to 30 m/s in 10 seconds.

Last edited: Jun 18, 2014
6. Jun 18, 2014

In the first 10 seconds, the initial velocity appears to be constant. In the second 10 seconds, the final velocity is 30 m/s and it sounds like it is constant then too. But the question presents the 20 m/s velocity to still be the initial velocity and I wasn't making the connection before on why the fact it is constant is important for this question, assuming I had the equation correct?

7. Jun 18, 2014

### ChiralWaltz

We can pick the appropriate equation once you get concept down. The question has two parts to it so we can't use that equation. I also don't recognize that equation but I am incredibly tired.

8. Jun 18, 2014

### haruspex

You can only use that equation for a period of the motion during which acceleration is constant. It is constant for the first 10 seconds, then constant for the second 10 seconds, but is not constant for the whole 20 seconds.
Calculate the distance covered in each 10 seconds separately.

9. Jun 18, 2014

### ChiralWaltz

What you are saying with your equation is that you are traveling 20 m/s and then somehow instantaneously traveling 30 m/s.

So put yourself in the car. What do you have to do to change your velocity from 20m/s to 30 m/s? What is this called in physics?

10. Jun 18, 2014

I understand the car accelerates. Before, I didn't understand the relevancy of finding acceleration with regard to the equation in my opening post.

For having a constant velocity, you will have an acceleration of zero, which is the case in the first 10 second interval and again in the second 10 second interval.

For dividing up each time interval and finding displacement:

Average velocity = change in displacement / change in time

For the first time interval:

Change in displacement = (20 m/s) (10 s) = 200 m

For the second time interval:

Change in displacement = (30 m/s) (10 s) = 300 m

I'm not sure where to go from here with it.

11. Jun 18, 2014

### bobie

This is the same problem of the electron you posed in the other thread, there it was t=S/V, here it is S=t*V .
10 s * 20 m/s makes 200 m
then next 10 second find the average speed, A*t, and you'll get 450

Last edited: Jun 18, 2014
12. Jun 18, 2014

### ChiralWaltz

I'm not sure you understand the car is accelerating.

I agree that a car with a constant velocity of 20 m/s goes 200 m in to seconds.

The velocity this car is traveling after 10 seconds is 20 m/s. from this initial velocity of 20 m/s, the car accelerates to a final velocity of 30 m/s. acceleration is a term used in physics to describe a change in velocity. The change in velocity occurs during the 20 m/s to 30 m/s. since acceleration is occurring, it is a non-zero value and you must take acceleration into account.

Last edited: Jun 18, 2014
13. Jun 18, 2014

### bobie

so from 0 to 10 s S=200m
from 10 to 20s, t= 10 * V (average speed) = ....m
200+ ....= 450 m.

14. Jun 19, 2014

### ChiralWaltz

You can do it that way also. I guess I was set on using a specific equation for some reason.