The position of a particle moving rectilinearly is given by `x = t^3 - 3 t^2 - 10`. Find the distance travelled by the particle in the first `4` seconds starting from `t = 0`.

Options

A. 24 unit
B. 36 unit
C. 15 unit
D. 50 unit

Answer

A

Explanation

`x = t^3 - 3t^2 - 10`
`v = (dx)/(dt) = 3t^2 - 6t`
Now, `v = 0` gives
`t = 0` and `t = 2 sec`
Velocity will becomes zero at `t =2 sec`., so particle will change direction after `t = 2 sec`.
At `t = 0`
`x_((0 sec)) = -10`
At `t = 2 sec`
`x_((2 sec)) = 2^3 - 3(2)^2 - 10 = 8 - 12 - 10 = -14`
At `t = 4 sec`
`x_((4 sec)) = 4^3 - 3(4)^2 - 10`
=`64 - 48 - 10 = 6`
Distance travelled `= x_1 + x_2`
=`|-14 -(-10)|+|6 -(-14)|=4 + 20 = 24`
Distance Travelled `= 24` units.
.