Question

$\huge \underline{\sf{✠ \: Question : - }}$


A car moving with a velocity of 10 m/s accelerates uniformly at 1 m/s^2 until it reaches a velocity of 15 m/s.

Calculate:–

(i) the time taken

(ii) the distance travelled during the acceleration.​

Answer 1

${\underline {\underline {\huge {\sf { Provided \: Question: } }}}}$

A car moving with a velocity of 10 m/s accelerates uniformly at 1 m/s² until it reaches a velocity of 15 m/s.

Calculate:–

(i) the time taken

(ii) the distance travelled during the acceleration.

${\underline {\underline {\huge {\sf { Required \: Answer: } }}}}$

${\underline {\boxed {\large {\sf {\pink { Time \: taken = 5sec} }}}}}$

${\underline {\boxed {\large {\sf {\pink { Distance \: travelled = 62.5m} }}}}}$

_____

Given:

• Initial velocity (u) = 10m/s

• final velocity (v) = 15m/s

• acceleration (a) = 1m/s²

To calculate:

• (i) the time taken
• (ii) the distance travelled during the acceleration.

Calculation:

${\underline {\boxed {\large {\sf {\blue { Calculating \: time \: taken \: (t) } }}}}}$

From the first equation of motion:

• ${\underline {\boxed {\large {\bf { v = u + at} }}}}$

Where,

$\implies$ v = final velocity

$\implies$ u = initial velocity

$\implies$ a = acceleration

$\implies$ t = time

Substituting the values to find time taken:

⠀⠀⠀

⠀⠀⠀$\bf { ⇢ 15 = 10 + ( 1 \times t ) }$

⠀⠀⠀

⠀⠀⠀

⠀⠀⠀$\bf { ⇢ 15 = 10 + t }$

⠀⠀⠀

⠀⠀⠀

⠀⠀⠀$\bf { ⇢ t = 15-10}$

⠀⠀⠀

⠀⠀⠀

⠀⠀⠀$\boxed { \bf { ⇢ t = 5 \: seconds } }$

Therefore, time taken was 5 seconds.

___________________________________

${\underline {\boxed {\large {\sf {\blue { Calculating \: distance \: travelled \: (s) } }}}}}$

From the 3rd equation of motion:

• ${\underline {\boxed {\large {\bf { 2as = {v}^{2} -{u}^{2} } }}}}$

Where,

$\implies$ v = final velocity

$\implies$ u = initial velocity

$\implies$ a = acceleration

$\implies$ s = distance travelled

Substituting the values to find distance travelled during the acceleration:–

⠀⠀⠀

⠀⠀⠀$\bf { ⇢ 2 \times 1 \times s = {15}^{2} - {10}^{2} }$

⠀⠀⠀

⠀⠀⠀

⠀⠀⠀$\bf { ⇢ 2s = 225-100}$

⠀⠀⠀

⠀⠀⠀

⠀⠀⠀$\bf { ⇢ 2s = 125 }$

⠀⠀⠀

⠀⠀⠀

⠀⠀⠀$\bf { ⇢ s = \dfrac{125}{2} }$

⠀⠀⠀

⠀⠀⠀

⠀⠀⠀$\boxed { \bf { ⇢ s = 62.5m} }$

Therefore, distance travelled during the acceleration is 62.5m.

$\rm \blue {So, we \: are \: done!! }$

___________________________________

$\huge \rm { ✭ More \: Information}$

Equations of motion:

• v = u + at
• s = ut + ½at²
• 2as = v² - u²

Laws of motion:

First law of motion:

✤ An object in motion will remain in motion unless acted upon by an outside force. An object at rest will remain at rest unless acted upon by an outside force.

Second law of motion:

✤ The acceleration of the body is directly proportional to the force acting on it and is inversely proportional to its mass and takes place in the same direction of the force.

Third law of motion:

✤For every action there is an equal opposite reaction.

__________________________________

Answer 2

Answer:-

Given:-

Initial Velocity of the car (u) = 10 m/s

Acceleration (a) = 1 m/s²

Final velocity (v) = 15 m/s.

We can find the Time taken by the car to

increase the velocity by using the first

equation of motion.

i.e.,

v = u + at

So, 15 = 10 + (1)(t)

⟹ 15 - 10 = t

⟹ 5 s = t

Now,

Using second equation of motion,

i.e.,

S = ut + 1/2 at²

Distance travelled by the car in this period of time is :

S = (10)(5) + 1/2 (1)(5)²

⟹ S = 50 + 25/2

⟹ S = 50 + 12.5

⟹ S = 62.5 m

∴

• Time taken by the car is 5 seconds.

• Distance travelled by the car is 62.5 m.