# Homework Help: Distance travel given constant a

1. Jul 8, 2010

### angelcase

A hot rod can accelerate from 0 to 60 km/h in 5.4 s.
(a) What is its average acceleration, in m/s2, during this time?

(b) How far will it travel during the 5.4 s, assuming its acceleration is constant?

(c) How much time would it require to go a distance of 0.32 km if its acceleration could be maintained at the value in (a)?

for a, first I did a unit conversion for the speed:

60(1000/3600)= 16.667m/s=v

a=delta v/ delta t

a=16.667/5.4=3.086m/s^2

for b:

d=V^2/2a
d=(16.667)^2/2(3.086)=45.006m

for c:

I converted .32km to 320m.

a=3.086m/s^2 since it is constant...but I don't know what equation to use to determine the time to travel the distance.

I tried the values 19.19 and 10.18, both were wrong.

2. Jul 8, 2010

### Karmalo

(a) and (b) results are correct.

If you put it like that it seems that 16.667 is dv and 5.4 is dt. You can do that because you can also say this: $$a_{m}=\frac{\Delta v}{\Delta t}$$ where $$\Delta$$ stands for increment, v for velocity and t for time. You can do this when the acceleration is constant.

For (c) just take the third kinematic equation.

Last edited: Jul 8, 2010
3. Jul 8, 2010

### Tusike

d=v^2 / (2a)

is equivalent to:

d=(a*t)^2 / (2a) = (a * t^2) / 2.
By the way, this I think is more commonly used.

4. Jul 8, 2010

### angelcase

Karmalo, what is the third kinematic equation?? I am horrible with physics and reading the book is doing no good...

5. Jul 8, 2010

### Karmalo

$$X=X_{0}+V_{0}t+\frac{1}{2}at^{2}$$
Have you ever used this one?

6. Jul 8, 2010

### angelcase

I don't know if I have...is x= to distance? In different books and sites they use different variable to represent distance, whether it be s, d or x.

if that is the case would I rearrange it to solve for t?

2(X-Xsub0)/(Vsub0+a)??

7. Jul 8, 2010

### Karmalo

$$X$$ is the final position and $$X_{0}$$ is the initial position so you can say $$X-X_{0}$$ is the distance traveled.

You can work out the value of the time (t) from the equation.

Note: $$V_{0}$$ is the initial speed

Last edited: Jul 8, 2010
8. Jul 8, 2010

### angelcase

I am not sure if I am doing this right.

I plugged in:

320-45-006=16.667t+ 1/2(3.086)t^2
for t I found it to be 3.886 then I added the 5.4sec already passed and got 9.286sec...does that sound correct, or am I not understanding.

9. Jul 8, 2010

### Karmalo

So we have this: $$X=X_{0}+V_{0}t+\frac{1}{2}at^{2}$$

We can pass $$X_{0}$$ to the left and we'd have $$X-X_{0}=V_{0}t+\frac{1}{2}at^{2}$$

The distance is 320m so $$X-X_{0}=320$$, we also know the initial speed is 0 so $$V_{0}=0$$ and the acceleration, which is 3,086 m/s^2

Now we just have to replace this values and we get this: $$320=0+\frac{1}{2}\cdot3,086t^{2}$$

$$t=\pm\sqrt{\frac{2\cdot320}{3,086}}$$

As time cannot be negative: $$t=14,4(s)$$