Question

AN OBJECT START'S FROM REST AND TRAVEL'S WITH AN ACCELERATION OF
5m/
$s {}^{2}$
WHAT WILL BE THE VELOCITY AFTER 3second?
​

Answer 1

Answer:

Explanation:

u=o

because initial object is starting from rest.

a=5m/s

v=?

t=3 sec

USING FIRST EQUATION OF MOTION

V=U+AT

V=0+5x3

V=15M/S

MARK IT AS BRAINLIEST

Answer 2

$\large{\underline{\rm{\green{\bf{Question:-}}}}}$

An object starting from rest travels with an acceleration of $\sf 5 \: m/s^{2}$. What will be the velocity after 3 seconds?

$\large{\underline{\rm{\green{\bf{Given:-}}}}}$

Acceleration = $\sf 5 \: m/s^{2}$

$\large{\underline{\rm{\green{\bf{To \: Find:-}}}}}$

The velocity after 3 seconds.

$\large{\underline{\rm{\green{\bf{Solution:-}}}}}$

We know that,

• u = Initial velocity of the particle
• A = Acceleration of the particle
• t = Time interval in which the particle is in consideration
• v = Velocity of the particle

Since the object is initially on rest, the initial velocity (u) will be zero.

According to the question,

Initial velocity (u) = 0

Acceleration (a) = $\sf 5 \: m/s^{2}$

Time (t) = 3 second

Now, finding the final velocity using the first equation of motion.

$\boxed{\sf v=u+at}$

$\longrightarrow \sf v=0+15 \times 3$

$\sf = \underline{\underline{v=15 \: m/s}}$

Therefore, the final velocity after 3 second will be 15 m/s.

$\large{\underline{\rm{\green{\bf{Note:-}}}}}$

When we are talking about motion in a straight line with constant acceleration, there are three equations of motion, which are helpful in determining one of the unknown parameters:

$\longrightarrow \sf v=u+at$

$\longrightarrow \sf x+ut+\dfrac{1}{2} \: at^{2}$

$\longrightarrow \sf v^{2}+u^{2}+2as$

One of the common examples of uniformly accelerated motion is freely falling bodies.