Question

A body has an initial velocity of 3 m/s and has an acceleration of 1 m/sec² normal to the direction of the initial velocity. Then, its velocity 4 seconds after the start, is:

Moderate

Solution

$\begin{array}{l} u_{x} = 3 m / s, a_{x} = 0 \\ u_{y} = 0, a_{y} = 1 m / \sec^{2} and t = 4 \sec \end{array}$
If v_x and v_y be the velocities after 4 sec respectively, then
$\begin{array}{l} v_{x} = u_{x} + a_{x} t = 3 {ms}^{- 1} \\ v_{y} = u_{y} + a_{y} t = 0 + 1 \times 4 = 4 {ms}^{- 1} \\ v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{3^{2} + 4^{2}} = \sqrt{9 + 16} = 5 m / s \end{array}$
Angle made by the resultant velocity w.r.t. direction of initial velocity, i. e.,x-axis, is
$β = \tan^{- 1} \frac{v_{y}}{v_{x}} = \tan^{- 1} (\frac{4}{3})$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

SCAN ME