Question

5. A particle is pushed along a horizontal surface in such a way that it starts with a velocity of 15 m/s. Its velocity decreases at a uniform rate of 0.25m/s2 (a) Find the time in which it will come to rest. (b) Find the distance covered by it before coming to rest? (2)​

Answer 1

Answer:

           i) t = 60 seconds

                ii) s = 450m

Explanation:

u = 15m/s

a = -0.25m/s²

i) v = 0

v = u + at

t = v - u / a

t = 0 - 15 / -0.25 = 15 / 0.25 = 1500/25

t = 60 seconds

ii) 2as = v² - u²

2as = -u²

s = -u² / 2a

s = - 15 × 15 / 2 × -0.25

s = 225 / 0.5

s = 450m