At a certain time a particle has a speed of 18 m/s in 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 the 2.4 s interval?

Question

At a certain time a particle has a speed of 18 m/s in 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 the 2.4 s interval? (A)  20 m/s2 (B)  10 m/s2 (C)  5 m/s2 (D)  2.5 m/s2

Tardigrade Admin · Accepted Answer

Average acceleration =( text change in veloclty / text Time taken ) =(-30 m / s -18 m / s /2.4)=20 m / s 2

Q. At a certain time a particle has a speed of $18 m / s$ in 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 the $2.4 s$ interval?

$20 m / s^{2}$

$10 m / s^{2}$

$5 m / s^{2}$

$2.5 m / s^{2}$

Average acceleration $= \frac{change in veloclty}{Time taken}$
$= \frac{- 30 m / s - 18 m / s}{2.4} = 20 m / s^{2}$

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

20m/s2

10m/s2

5m/s2

2.5m/s2