Using calculus to solve motion in straight line

Question

Suppose a certain car supplies a constant deceleration of A meters per second per second. If it is traveling at 90 kilometers per hour (25 meters per second) when the brakes are applied, its stopping distance is 50 meters.

(a) What is A?

b) What would the stopping distance have been if the car had been traveling at only 54 kilometers per hour when the brakes were applied?

(c) At what speed is the car traveling when the brakes are applied if the stopping distance is 56 meters?

Answer 1

(a) A = 6.25 m/sec², derivation below followed by answers to the other parts.

Let s(t) = distance of car from where it was at the point in time when brakes first applied (i.e., t=0 is the moment at which the brakes are applied.) We're setting up the problem so that, by definition, s(0) = 0.

We know s''(t) = -A (negative because the car is slowing down).

Therefore s'(t) = -At + C and s(t) = -At²/2 + Ct +D where C and D are constants of integration. Since s(0) = 0, we know D is equal to 0.

On the other hand, we're told that s'(0) = 25 m/sec. so 25 = -A*0 + C = 25

At time t = t_STOP, the car has traveled 50 m and its derivative is 0.

0 = s'(t_STOP) = -At_STOP + 25, or t_STOP = 25/A

Therefore 50 = s(t_STOP) = -At²/2 + 25t.

Plug in 25/A for t_STOP to get 50 = (-A/2)*(25/A)² - 25(25/A). Solving (just algebra, no more calculus necessary at this point), 50 = 625/2A and so A = 6.25

(b) In this case t_STOP = 15/6.25 = 2.4 secs (see below for the reasoning).

Therefore the stopping distance s(t_STOP) = -6.25/2 (2.4)² + 15 (2.4) = 18 ft

54 km/hr = 54,000 meters / 3600 seconds = 15 m/sec

In this case we have 0 = s'((t_STOP) = -6.25t_STOP + 15 (here I'm assuming that the deceleration is the same as in part (a), but have inserted 15 m/sec for initial speed).

For (c), the idea is similar.