# Velocity Question

1. Oct 6, 2009

### 1irishman

1. The problem statement, all variables and given/known data
A car of length 4m and a bus of length 12m are travelling in the same direction 20m apart. The cars current velocity is 10m/s and the buses current velocity is 20m/s, the car suddenly starts to accelerate at 5m/s^2 and stops accelerating after 5 seconds. How long does it take the car to completely pass the bus?

2. Relevant equations
d=Vit + 1/2at^2
i'm not sure about other ones that can be used here.....

3. The attempt at a solution
i figured (not sure if it's right though) the bus travels 100m in 5s and the car travels 112.5m in the same amount of time. So the original distance of 20m between the car and the bus has lessened to 20m - 12.5m = 7.5m. So the distance between the bus and the car is now 7.5m. Not sure how to proceed from here...hints? Thanks.

Last edited: Oct 6, 2009
2. Oct 6, 2009

### rl.bhat

What is the velocity of the car after 5 seconds?
Problem says that the car and bus are 20 m part. But it is not clear form which point to which point

3. Oct 6, 2009

### 1irishman

The velocity of the car after 5 seconds is 10m/s (back to its original velocity).

4. Oct 6, 2009

### rl.bhat

This is not correct. Use
v = vo + at

5. Oct 6, 2009

### 1irishman

are my distances wrong then?

6. Oct 6, 2009

### rl.bhat

They are correct.

7. Oct 6, 2009

### 1irishman

Okay i have:
if vi of car=10m/s
a=5m/s^2
t=5s
then vf=35m/s
Is that right so far? thanks.

8. Oct 6, 2009

### rl.bhat

Yes. It is correct.

9. Oct 6, 2009

### 1irishman

okay...i think i've solved it:
since cars new constant speed is 35m/s and distance is 23.5m.
The relative speed to the bus should be 15m/s.
So, v=15m/s
d=23.5m
then t=1.6s
So, it should take the car 6.6s to pass the bus right?

10. Oct 6, 2009

### rl.bhat

v = d/t.
Then t = ?

11. Oct 6, 2009

1.6s plus 5s