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Standard IX

Physics

Velocity(Speed with Direction)

A particle st...

Question

A particle starts moving towards positive $x -$ axis with initial velocity $4 m / s$ . Acceleration of the particle is $2 m / s^{2}$ towards negative $x -$ axis. Find the distance (in $m$ ) travelled by the particle in $4 s$ .

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Solution

Given,
Initial Velocity $= 4 m / s$
Acceleration $= - 2 m / s^{2}$
$t = 4 s$
Now,
$v = u + a t$
$v = u - 2 t$
At $t = 2 s, v = 0$
For $t = 0$ to $t = 2$ , particle moves along $+ v e x -$ direction and for $t = 2$ to $t = 4$ , particle moves along $- v e x -$ direction.
So, total distance travelled,
$v^{2} - u^{2} = 2 a s$
$s = \frac{(4^{2})}{2 \times 2} = 4 m$
total distance travelled $= 2 s = 8 m$

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Velocity(Speed with Direction)

Standard IX Physics

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A particle starts moving towards positive x − axis with initial velocity 4 m/s. Acceleration of the particle is 2 m/s2 towards negative x − axis. Find the distance (in m) travelled by the particle in 4 s.

A particle starts moving towards positive $x -$ axis with initial velocity $4 m / s$ . Acceleration of the particle is $2 m / s^{2}$ towards negative $x -$ axis. Find the distance (in $m$ ) travelled by the particle in $4 s$ .