Question

A body with an initial velocity of 4m/s accelerates uniformly at the rate of 1.2 m/s² over a distance of 20 m. Calculate: (i) the final velocity of the body.(ii) the time in which the body covers that distance. (iii) the average velocity of the body during the journey

Answer 1

Given :

1. Initial velocity of body = 4m/s
2. Accelaration = 1.2 m/s²
3. Distance covered = 20m

To find :

• Final velocity of the body
• The time in which the body covers that distance.
• The average velocity og te body during the journey.

${\purple{\boxed{\large{\bold{Formula's}}}}}$

Kinematic equations for uniformly accelerated motion .

$\bf\:v=u+at$

$\bf\:s=ut+\frac{1}{2}at{}^{2}$

$\bf\:v{}^{2}=u{}^{2}+2as$

and $\bf\:s_{nth}=u+\frac{a}{2}(2n-1)$

Solution :

1) We have to find the final velocity of the body .

By equation of motion

$\sf\:v{}^{2}=u{}^{2}+2as$

$\sf\:v{}^{2}=4^2+2\times1.2\times20$

$\sf\:v{}^{2}=16+48$

$\sf\:v{}^{2}=64$

$\sf\:v{}^{2}=\sqrt{64}$

$\sf\:v=8ms^{-1}$

Thus , the final velocity of the body is 8 m/s

2) We have to find the time in which the body covers that distance

By equation of motion

$\sf\:v=u+at$

$\sf\:8=4+1.2\times\:t$

$\sf\:8-4=1.2t$

$\sf\:1.2t=4$

$\sf\:t=\dfrac{4}{1.2}$

$\sf\:t=3.33\:sec$

Thus , The time in which the body covers that distance is 3.33 sec

3) We have find Average velocity of the body During this time .

We know :

$\sf\:\blue{Average\:Velocity=\dfrac{Displacement}{Time}}$

$\sf\:Average\:Velocity=\dfrac{20}{3.33}$

$\sf\:Average\:Velocity=6.006ms^{-1}$

Thus , Average velocity of the body During this time is 6.006 m/s.

Answer 2

Answer:

Given :

Initial velocity of body = 4m/s

Accelaration = 1.2 m/s²

Distance covered = 20m

To find :

Final velocity of the body

The time in which the body covers that distance.

The average velocity og te body during the journey.

{\purple{\boxed{\large{\bold{Formula's}}}}}

Formula

′

s

Kinematic equations for uniformly accelerated motion .

\bf\:v=u+atv=u+at

\bf\:s=ut+\frac{1}{2}at{}^{2}s=ut+

2

1

at

2

\bf\:v{}^{2}=u{}^{2}+2asv

2

=u

2

+2as

and \bf\:s_{nth}=u+\frac{a}{2}(2n-1)s

nth

=u+

2

a

(2n−1)

Solution :

1) We have to find the final velocity of the body .

By equation of motion

\sf\:v{}^{2}=u{}^{2}+2asv

2

=u

2

+2as

\sf\:v{}^{2}=4^2+2\times1.2\times20v

2

=4

2

+2×1.2×20

\sf\:v{}^{2}=16+48v

2

=16+48

\sf\:v{}^{2}=64v

2

=64

\sf\:v{}^{2}=\sqrt{64}v

2

=

64

\sf\:v=8ms^{-1}v=8ms

−1

Thus , the final velocity of the body is 8 m/s

2) We have to find the time in which the body covers that distance

By equation of motion

\sf\:v=u+atv=u+at

\sf\:8=4+1.2\times\:t8=4+1.2×t

\sf\:8-4=1.2t8−4=1.2t

\sf\:1.2t=41.2t=4

\sf\:t=\dfrac{4}{1.2}t=

1.2

4

\sf\:t=3.33\:sect=3.33sec

Thus , The time in which the body covers that distance is 3.33 sec

3) We have find Average velocity of the body During this time .

We know :

\sf\:\blue{Average\:Velocity=\dfrac{Displacement}{Time}}AverageVelocity=

Time

Displacement

\sf\:Average\:Velocity=\dfrac{20}{3.33}AverageVelocity=

3.33

20

\sf\:Average\:Velocity=6.006ms^{-1}AverageVelocity=6.006ms

−1

Thus , Average velocity of the body During this time is 6.006 m/s.