Question

A Car is initially at rest starts moving with a uniform acceleration of 0.5m/s² and moves with it for 5 minutes find its velocity after 5 min and and distance Traveled in 5 mins .​

Answer 1

Answer :-

→ Velocity after 5 mins is 150 m/s .

→ Distance travelled is 22.5 km .

Explanation :-

We have :-

• Initial velocity (u) = 0 m/s

• Acceleration (a) = 0.5 m/s²

• Time taken (t) = 5 mins = 300 sec

_____________________________

Firstly, let's calculate the velocity acquired by the car in given time (5 min) by using the 1st equation of motion .

v = u + at

⇒ v = 0 + 0.5(300)

⇒ v = 0 + 150

⇒ v = 150 m/s

Now, we can calculate the distance travelled by the car by using the 2nd or 3rd equation of motion. Here, let's do with 2nd equation .

s = ut + ½at²

⇒ s = 0(300) + ½ × 0.5 × (300)²

⇒ s = 0 + ½ × 0.5 × 90000

⇒ s = 0 + 0.5 × 45000

⇒ s = 22500 m

⇒ s = 22.5 km

Answer 2

Given:

Initial velocity = 0m/s

Acceleration = 0.5m/s^2

Time taken = 5minutes = 300seconds

To find:

• The velocity

• The distance travelled

Solution:

★ Calculating velocity

Using 1st equation of motion

v = u + at

Where,

v is velocity

u is initial velocity

a is acceleration

t is time

☯︎ Putting values in the formula

v = 0 +0.5 (300)

v = 0 + 150

v = 150m/s

★ Calculating distance travelled

Using second equation of motion

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

☯︎ Putting values in the formula

$\sf\rightsquigarrow \: \: {s=0(300) + \frac{1}{2} \times 0.5 \times (300)^{2}}$

$\sf\rightsquigarrow \: \:{s = 0 + \frac{1}{2} \times 0.5 \times 90000 }$

$\sf\rightsquigarrow \: \:{s = 0 + 0.5 \times 45000}$

$\sf\rightsquigarrow \: \:{s = 22500m}$

$\sf\rightsquigarrow \: \:{s = 22.5km}$

Therefore,

• The velocity is 150m/s

• The distance travelled is 22.5km