# Car physics homework help

1. Jan 22, 2007

### hemang

Two cars, A and B, are traveling in the same direction, although car A is 208 m behind car B. The speed of A is 25.7 m/s, and the speed of B is 18.6 m/s. How much time does it take for A to catch B?
Time it takes for A to catch B is
speed=distance/time
time=distance/speed
=208/25.7
=8.09s

Is that right.
if not, help me
thanks

2. Jan 22, 2007

### cristo

Staff Emeritus
No, you're not correct. If you want to use that formula, you need to consider the difference between the speeds of the two cars, not the speed of car A.

3. Jan 22, 2007

### hemang

I don't get it. could u describe clearly?

4. Jan 22, 2007

### hemang

car A: time=(d-208)/25.7
so, d=(25.7m/s)t + 208m
car B: time=(d+208)/18.6
so, d=(18.6m/s)t-208

(25.7m/s)t + 208m = (18.6m/s)t - 208m
(7.10m/s)t = -416m
t= (-416m)/(7.10m/s)
t=58.6 seconds
Is this right?

5. Jan 22, 2007

### cristo

Staff Emeritus
No, this is not correct, but you're getting closer! Look at the distances you have assigned to the cars. According to your equations, how far behind B is car A at t=0?

We want to find the time when the displacements of the two cars are equal. Let's say that at a distance d from car B's initial position, car A catches it up. How far does car A have to travel to reach this point? This should help you set up your equations.

6. Jan 22, 2007

### hemang

so, would that be 208-d?

7. Jan 22, 2007

### hemang

t for A: (208m-d)/25.7m/s
so, d= 208m-(25.7m/s)t
t for B: (208+d)/18.6m/s
so, d= (18.6m/s)t-208m

(18.6m/s)t-208m = 208m- (25.7m/s)t
t=416m/44.3m/s
=9.39 seconds

8. Jan 22, 2007

### hemang

Two cars, A and B, are traveling in the same direction, although car A is 208 m behind car B. The speed of A is 25.7 m/s, and the speed of B is 18.6 m/s. How much time does it take for A to catch B?

help me with equations.

9. Jan 22, 2007

### cristo

Staff Emeritus
There's no need to repost the question.

No, car A need to travel more distance.. so the total distance will be 208+d.

You didn't listen to what I said. Again, when t=0, the distance between the two cars in your equations is 416m.. this is not correct.

Let the distance B covers be equal to d. Then, car A will have to travel 208+d to reach the same point. Try writing equations with these values for distance.

10. Jan 22, 2007

### hemang

thanks. i got it.
(25.7m/s)t=(18.6m/s)t-208
t=208m/(7.10m/s)
t=29.3 seconds