Two cars are traveling along a straight-line in the same direction, the lead car

Question

Two cars are traveling along a straight-line in the same direction, the lead car at 25.0 m/s and the other car at 35.0 m/s. At the moment the cars are 40.0 m apart, the lead driver applies the brakes, causing his car to have an acceleration of -2.10 m/s2.

(a) How long does it take for the lead car to stop? 11.9 seconds

(b) Assuming that the chasing car brakes at the same time as the lead car, what must be the chasing car's minimum negative acceleration so as not to hit the lead car?

(c) How long does it take for the chasing car to stop?

I know (a) is 11.9 seconds, but I can't figure out which equation to use for (b). Also, I need (b) to solve for (c) if I'm not mistaken.

Explanation / Answer

Your calculation of 11.9s for (a) is correct.

b) To understand how to do this part, we need to break the question down into the terms of motion. Treating the cars as simple points, the cars will hit when they reach the same position, so we first need to determine the position at which the lead car stops, and then determine the (negative) acceleration required by the second car to stop at the same position.

To find the position the leading car stops at, we can use the formula (the Cramster equation editor isn't working for me for some reason):

v^2 = v_i^2 + 2a(x-x_i)

If this formula isn't given for you, you can derive it from

v = v_i + at

x = x_i + v_i*t + 0.5at^2

Solve for t in the first formula and plug it into the second.

Solving for x and taking final velocity 0, we get

x = -v_i^2/(2a)+x_i

Plugging in the numbers (we will take the initial position of the trailing car as 0m):

x = -(25.0m/s)^2/(2*(-2.10 m/s^2))+40.0m=189m

Now we use the same formula for the trailing car, but solve for a instead. Once again, final velocity is 0, and we also assume initial position is 0:

a = -v_i^2/(2x)

Plugging in the initial velocity and same value of x that we just calculated:

a = -(35.0m/s)^2/(2*189m)=-3.24 m/s^2

c) Same as part (a), using the acceleration we just computed.

v = v_i + at

Solve for t, final v=0:

t = -v_i/a = -35.0m/s/-3.24m/s^2=10.8s

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