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A motorist passes a school-crossing corner where the speed limit is 10m/s, travelling with a constant velocity of 15m/s. So a police officer on a motor cycle at the corner starts off in pursuit with constant acceleration of 3m/s²

a) How much time elapses before the officer catches up with the car?

b) What is the officer's speed at this point?

c) What is the total distance the officer has travelled at this point?

ghanamichael247:
I need help with question 1

type the question, I'd have to stress my eyes to see from the exercise book

You should type the questions and make sure you do it correctly

speed limit of the school (v) = 10m/s constant velocity (u) = 15m/s constant acceleration = 3m/s^2
eqns of motion 1. a = v-u/t 2. v^2 = u^2 + 2as 3. s = ut + 1/2at^2.........
a.time a = v-u/t 15-10/3 = 1.67seconds
b. speed speed = distance/time, v^2 = u^2 + 2as 15^2 = 10^2 + 2*3*s 225-100 = 6s s = 125/6= 20.83m but speed = distance/time speed = 20.83/1.67 = 12.47m/s.
c. total distance s = ut + 1/2at^2 s = 15*1.67 + 1/2*3*1.67^2 s = 29.23m total distance = 29.23+20.83 = 50.06m.

5s,15m/s and 37.5m.

The officer started his chase from rest, so initial velocity,u is 0.
He aims to catch up with the driver at his speed i.e
v
Neglecting the speed limit i.e 10m/s, and applying the equation of uniformly accelerated motion.

Hope i am correct. It's been long since i last solved this kind of physics.

Go to university physics and copy d answer joor.