Question

a man across a 90 m long straight track with a uniform acceleration in 6 s.if his initial velocity is 3m/s, then he leaves the track with velocity

Answer 1

Answer on Question#49232 - Physics - Mechanics - Kinematics

A man across a 90 m long straight track with a uniform acceleration in 6 s.if his initial velocity is $3\mathrm{m/s}$, then he leaves the track with velocity.

Solution:

The displacement of the body moving with constant acceleration can be given by

$$s = \frac{v_0 + v}{2} t$$

where $s$ is the displacement, $t$ is the time, $v_0$ is the initial velocity, $v$ is the final velocity. Then the final velocity can be expressed as follows:

$$v = \frac{2s}{t} - v_0$$

Substituting $s = 90 \, \text{m}$, $t = 6 \, \text{s}$, $v_0 = 3 \, \frac{\text{m}}{\text{s}}$ we obtain $v = 27 \, \frac{\text{m}}{\text{s}}$.

**Answer:** 27 m/s.

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