Question

A sphere is fired downward into a medium with an initial speed of 27 m/s. If it experiences a deceleration $\ddot{x}=(-6t)m/s^2$, where t is in seconds. Determine the distance travelled before it stops.(in m)

• A. 45
• B. 54
• C. 27
• D. 81

Answer 1

The correct option is B 54

Acceleration $\ddot{x}=-6t$

$i.e.,\frac{d\dot{x}}{dt}=-6t$

$d\dot{x}=-6tdt$

Integrating both sides

${\int }_{27}^{\dot{x}}d\dot{x}={\int }_0^t-6tdt$

(Given $\dot{x}=27$m/s when t = 0)

$\dot{x}-27=-3t^2$

$\dot{x}=27-3t^2\text{(Velocity-time relationship)}$

$i.e.,\frac{dx}{dt}=27-3t^2$

$dx=(27-3t^2)dt$

Integrating both sides assuming the point of projection as the origin

${\int }_0^{x}dx={\int }_0^t(27-3t)dt$

$x=27t-t^3\text{(displacement-time relationship)}$

Here we have to find out the distance travelled before it comes to rest.

When the sphere comes to rest, $\dot{x}=0$

$i.e.,27-3t^2=0$

t = 3 seconds

$\therefore$ displacements after 3 seconds

$x=27\times 3-3^3=54m$