At a certain time a particle had a speed of 18 m/s in the positive x direction, and 2.4 s later its speed was 30 m/s in the opposite direction. What is the magnitude of the average acceleration of the particle during this 2.4 s interval?

m/s2

a) in the initial direction of motion

b) opposite the initial direction of motion

c) direction changes continuously

(a) (18m/s/t₁) m/s²

(b) - 12.5m/s²

(c) - 20mls²

Explanation:

(a) Let t₁ be the initial time

a = v-u/t

acc = (18m/s/t₁) m/s²

(b) acc = - 30m/s/2.4

= - 12.5m/s²

(c) The particle was at a speed of 18m/s in the positive x-direction and later after 2.4s ≡Δt, it was at speed of - 30m/s in the negative x-direction.

so this imply that the velocity was first v₁ = 18m/s and later v₂ = - 30m/s.

The average acceleration is then:

Aavg = Δv

Δt

= v₂-v₁/Δt

= - 30-18/2.4 = - 20mls²