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1st Equation of Motion

The accelerat...

Question

The acceleration of the particle moving in positive $x -$ direction with an initial velocity of $10 m/s$ is $3 {m/s}^{2}$ . Its velocity at $t = 4 s e c$ is

20 m/s

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22 m/s

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24 m/s

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18 m/s

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Solution

The correct option is B $22 m/s$
Given, $a = 3 {m/s}^{2}, u = 10 m/s and t = 4 s e c$
Using first equation of motion we have
$v = u + a t$
$\Rightarrow v = 10 + 3 \times 4 = 22 m/s$
Hence, velocity at $t = 4 s e c$ is $22 m/s$ .

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Similar questions

x -

3 m/s

2 {m / s}^{2}

t = 3 s

x = 4 t^{2} + 2 t

m

2 s

2 {m/s}^{2}

10 m/s

4 s

v = 3^i + 4^j m / s

0.4^i + 0.3^j m / s^{2}

10 s

a (m / s^{2}) = 3 t^{2} + 1

t = 1

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