The acceleration of the particle moving in positive $x -$ direction with an initial velocity of $20 m/s$ is $4 {m/s}^{2}$ . Its velocity at $t = 5 sec$ is

22 m/s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

40 m/s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

30 m/s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

25 m/s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $40 m/s$
Given, $a = 4 {m/s}^{2}, u = 20 m/s and t = 5 sec$
Using first equation of motion we have
$v = u + a t$
$\Rightarrow v = 20 + 4 \times 5 = 40 m/s$
Hence, velocity at $t = 5 sec$ is $40 m/s$ .

Suggest Corrections

Similar questions

x -

6 m / s

4 m / s^{2}

t = 5 s e c

40 m

5 s e c

65 m

5 s e c

A particle moves in a straight line with velocity 20 $\frac{m}{s}$ and acceleration -4 $\frac{m}{s^{2}}$ . Find the velocity after 5 seconds.

v (t) = (4 t + 6) m/s

10 sec

(a - s)

30 m

10 m/s

Join BYJU'S Learning Program