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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

Respuesta :

greenkelvin5

Answer:

(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

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