Question

Any body start from rest and accelerated with acceleration 2 m/sec^2 for 10 sec. Find its final velocity and distance moved in this interval.​

Answer 1

Answer:

The initial velocity travelled is 625m

The accleration is α=10m/s

2

The time is t=5s

Apply the equation of motion

v=u+at

The velocity is v=0+10.5=50m/s.

The distance travelled in the first 5s is

s=ut+

2

1

at

2

=0.5+

2

1

.10.5

2

=125m

Thedistancetravelledinthefollowing10sis

s

1

=ut

1

=50.10=500m

Thetotaldistancetravelledis

d=s+s

1

=125+500=625m

Answer 2

$\textbf{Solution:}$

Given:

u(initial velocity) =0 m/s

a(acceleration) = 2 m/s^2

t(time) =10 sec

To find:

• v(final velocity) and s(distance)

Applying first equation of motion:

$= > v = u + at$

$= > v = 0 + 2.10$

$= > v = 20 \: m {s}^{ - 1}$

Now, apply second equation of motion:

$= > s = ut + \frac{1}{2} a {t}^{2}$

$= > 0 \times 10 + \frac{1}{2} \times 2 \times {10}^{2}$

$= > s = 100 \: m$

Answer:

Therefore, v(final velocity) = 20 m/s

and s(distance travelled ) =100 m