Question

A particle is moving along a straight line through a fluid medium such that its speed is measured as v=(2t) m/s, where t is in seconds. If it is released from rest at s=0, determine its positions and acceleration when t=3 s.

Answer 1

Answer:

2 m/s², 19.62 m

Explanation:

v = Final velocity = 2t

t = Time taken

u = Initial velocity

s = Displacement

a = Acceleration

From the equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{2t-0}{t}\\\Rightarrow a=2\ m/s^2$

The acceleration of the particle is 2 m/s²

at t = 3 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 2\times 3^2\\\Rightarrow s=19.62\ m$

The position at t = 3 seconds is 19.62 m