Question

Two children are playing with a pair of wind-up toy cars. The children sit 2.5 meters apart and release
the two cars from rest heading towards each other on a collision course. One car accelerates at a rate
of 2.5 cm/s and the other car accelerates at a rate of 3.5 cm/s. How long after they are released will
the two cars collide?
(a) 0.91 s
(b) 1.6 s
(c) 2.5 s
(d) 6.5 s
(e) 9.1 s

Answer 1

I feel like sometime is wrong with this problem whether it be velocity or the distance between the cars. I just can not figure out this problwm

Answer 2

Well i can see one thing. 2.5 and 3.5 need to be m/s^2

Answer 3

Thanks let me see if I can go from that

Answer 4

yeah i can't get it still haha

Answer 5

i got like 0.8 sec

Answer 6

i will retry

Answer 7

i know the answer is 9.1 but i just cant figure out

Answer 8

I would do this:

X(t)= 2.5X10^-2 t^2 /2 plug in t ---->1.035

then

X(t)= 3.5X10^-2 t^2 /2 -----> 1.449

add 1.4 and 1.03 to get 2.5

if you plug in other values you won't get 2.5 when you add them.

Answer 9

okay so i plugged that in for my r(final position) vector and I'm not getting 9.1. Are you?

Answer 10

plug in 9.1 for t into my equations.

Answer 11

@jjpetty99
For this you need to use a reference frame. pretend one car is stationary. From your point of view (being still) the other car accelerates towards you at 0.06m/s^2. The time it takes to get to you from 2.5m away is the same as the time if you were both travelling towards each other. Fellowroot's equation was the right one, do:
s = ut + 0.5at^2
s = 2.5, u = 0, a = 0.06
Solve for t... what do you get? ;)