Question

A car traveling at 30 m/s runs a stop sign next to a stationary cop car. The cop accelerates at a rate of 3.2 m/s^2. Assume the car travels at the same rate.
(a) How long does the cop car take to catch up to the car?
(b) How far do the cars travel?
(c) What is the final speed of the cop car?
(d) Graph the position, velocity, and acceleration of both cars. For each graph, place both cars on the same graph (e.g. draw the position of the car and cop car on the same graph)

Answer 1

sorry - i got called away

Answer 2

a) the time taken by cop and distance travelled by the cop and those of the other car are the same

for other car
s = 30 t where s = distance
for cop car
s = 0.5*3.2*t^2 using constant acceleration equation

so 30t = 1.6t^2
1.6t^2 - 30t = 0
t(1.6t - 30) = 0
t = 18.75 seconds

Answer 3

(b) distance travelled = 30 * 18,75 = 562.5 m

Answer 4

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use v = at for cop car

final speed = 3.2 * 18.75 = 60 m/s

Answer 5

@cwrw238 how would the graph look like?

Answer 6

draw a velocity/time graph. velocity as vertical axis

graph of other car will be a straight parallel to horizontal axis through v=30

Answer 7

graph of cop's car will pass through origin and have a slope of 3.2 (straight line) top value of v = 60