Aspeeding car is travelling at a constant 30.0 m/s when it passes a stationary police car. if the police car delays for 1.00 s before starting, what must be the magnitude of the constant acceleration of the police car to catch the speeding car after the police car travels a distance of 300 m?
(a) 6.00 m/s2
(b) 3.00 m/s2
(c) 7.41 m/s2
(d) 1.45 m/s2
(e) 3.70 m/s2

option (C)

Explanation:

Speed of car = 30 m/s

Let the time taken by the police car to catch the speeding car is t

The distance traveled by the speeding car in t + 1 second is equal to the distance traveled by the police car in time t

Distance traveled by the police car in time t

   .... (1)

Distance traveled by the speeding car in t + 1 second

s = 30 (t + 1) = 300

t + 1 = 10

t = 9 s

Put the value of t in equation (1), we get

300 = 0 + 0.5 x a x 9 x 9

a = 7.41 m/s^2

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.