Question

A particle, whose acceleration is constant, is moving in the negative x direction at a speed of 4.91 m/s, and 12.9 s later the particle is moving in the positive x direction at a speed of 7.12 m/s.What is the particle’s velocity, in m/s, 12.4 s before it was moving in the negative x direction at a speed of 5.32 m/s?

Answer 1

Answer:

The particle’s velocity is -16.9 m/s.

Explanation:

Given that,

Initial velocity of particle in negative x direction= 4.91 m/s

Time = 12.9 s

Final velocity of particle in positive x direction= 7.12 m/s

Before 12.4 sec,

Velocity of particle in negative x direction= 5.32 m/s

We need to calculate the acceleration

Using equation of motion

$v = u+at$

$a=\dfrac{v-u}{t}$

Where, v = final velocity

u = initial velocity

t = time

Put the value into the equation

$a=\dfrac{7.12-(-4.91)}{12.9}$

$a=0.933\ m/s^2$

We need to calculate the initial speed of the particle

Using equation of motion again

$v=u+at$

$u=v-at$

Put the value into the formula

$u=-5.321-0.933\times12.4$

$u=-16.9\ m/s$

Hence, The particle’s velocity is -16.9 m/s.