Question

Velocity of a body increases at a constant rate 100 ms-1 to 500 ms-1 in 20 seconds. Find the acceleration:

Answer 1

Given:

• Initial velocity of an object was 100m/s.
• Final velocity of an object was 500m/s.
• Time taken to accelerate is 20 seconds.

To Find:

• The acceleration of the body.

Answer:

Given to us is that a body having a initial velocity of 100m/s increases its velocity to 500m/s taking a time of 20 seconds.

So , we know acceleration as ,

• The rate of change of velocity is called acceleration.

Formula is:

$\large\boxed{\red{\bf{\leadsto acceleration=\dfrac{v-u}{t}=\dfrac{\Delta v}{\Delta t}}}}$

where

• u is initial velocity.
• v is final velocity.
• t is time

$\sf{\implies accel^{n}= \dfrac{500 m s^{-1}-100 m s^{-1}}{20 s}}$

$\sf{\implies accel^{n}=\cancel{\dfrac{400 m s^{-1}}{20s}}}$

${\underline{\boxed{\red{\sf{\longmapsto accel^{n}= 20m s^{-2}}}}}}$

Hence the acceleration of the object was 20m/s².

Answer 2

Answer:

Given:

Initial velocity of an object was 100m/s.

Final velocity of an object was 500m/s.

Time taken to accelerate is 20 seconds.

To Find:

The acceleration of the body.

Answer:

Given to us is that a body having a initial velocity of 100m/s increases its velocity to 500m/s taking a time of 20 seconds.

So , we know acceleration as ,

The rate of change of velocity is called acceleration.

Formula is:

\large\boxed{\red{\bf{\leadsto acceleration=\dfrac{v-u}{t}=\dfrac{\Delta v}{\Delta t}}}}

⇝acceleration=

t

v−u

=

Δt

Δv

where

u is initial velocity.

v is final velocity.

t is time

\sf{\implies accel^{n}= \dfrac{500 m s^{-1}-100 m s^{-1}}{20 s}}⟹accel

n

=

20s

500ms

−1

−100ms

−1

\sf{\implies accel^{n}=\cancel{\dfrac{400 m s^{-1}}{20s}}}⟹accel

n

=

20s

400ms

−1

{\underline{\boxed{\red{\sf{\longmapsto accel^{n}= 20m s^{-2}}}}}}

⟼accel

n

=20ms

−2

Hence the acceleration of the object was 20m/s².