Question

A body has an inirial velocity of 3m/sec and an acceleration of 1m/sec^2 normal to the direction of initial velocity. then the velocity of the body 4 seconds after start is

Answer 1

Hii friend,

Here is your answer.....

By using the formula of equation of motion,

v =u +at

V = 3 + (1×4)

v =3+4

final velocity (V) =7 m/s.

Hope this helps you a little!!!

Answer 2

$\maltese \: \red{\mid{\underline{\overline{\textbf{Answer}}}\mid}}$

Velocity of body after 4s = 7m/s

$\maltese \:\underline \red{\mid{\underline{\overline{\textbf{Explanation :}}}\mid}}$

Acceleration of a body is the rate of change of velocity of a body. It is defined by the expression :

$\maltese \: a = \frac{v - u}{t}$

In this expression :

• a = acceleration
• v = final velocity
• u = initial velocity
• t = time take to accelerate

Now after shifting T to L. H. S we get :

$\maltese \: at = v - u \\ \\ \maltese \: u + at = v$

Therefore we got that final velocity of a body is sum of its initial velocity and product of acceleration and time.

$\maltese \: Coming \: to \: question :-$

Now we have been given the initial velocity of body and its acceleration we have to find its velocity after 4seconds.

Based on the above data we say :-

• Final velocity = ?
• Initial velocity = 3m/s
• Acceleration = 1m/s^2
• Time = 4seconds.

After inputting the values in derived equation we get :

$\maltese \: v = 3 + (1 \times 4)$

$\maltese \: v = 3 + 4$

$\maltese \: v = 7 \frac{m}{s}$

Therefore final velocity of body is 7m/s.

______________________________________

BY SCIVIBHANSHU

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