Question

A car accelerates uniformly from 20 km per hour to 40 km per hour in 10 second calculate its acceleration and the distance covered by the car in that time

Answer 1

Step-by-step explanation:

Given:-

• Initial velocity (u) = 20km/h

• Final velocity (v) = 40km/h

• Time (t) = 10sec

To Find:-

• Acceleration (a) = ?

• Distance (s) = ?

Solution:-

To convert km/h to m/s multiply it with 5/18.

Converting 20km/h to m/s

$\sf = 20\times \dfrac{5}{18}$

$\sf = 5.55$

Initial velocity (u) = 5.55m/s

Converting 40km/h to m/s

$\sf = 40\times \dfrac{5}{18}$

$\sf = 11.11$

Final velocity (v) = 11.11m/s

Acceleration:-

According to the first equation of motion:-

$\sf \dashrightarrow v = u + at$

$\sf \implies 11.11 = 5.55 + a( 10 )$

$\sf \implies 11.11 - 5.55 = 10a$

$\sf \implies 5.55 = 10a$

$\sf \implies a = \dfrac{5.55}{10}$

$\sf \implies a = 0.55$

$\underline{\boxed{\pink{\rm \therefore Acceleration = 0.55m/{s}^{2}}}}$

Distance covered by the car:-

According to the third equation of motion:-

$\sf \dashrightarrow s = ut + \dfrac{1}{2} a {t}^{2}$

$\sf \implies s = 5.55 \times 10 + \dfrac{1}{2} \times 0.55 \times {(10)}^{2}$

$\sf \implies s = 55.5 + 0.278 \times 100$

$\sf \implies s = 55.5 + 27.8$

$\sf \implies s = 83.3$

$\underline{\boxed{\purple{\rm \therefore Distance = 83.3m}}}$

Answer 2

Given :

• Initial velocity (u) = 20 km/hr = 5.55 m/s

• Final velocity (v) = 40 km/hr = 11.11 m/s

• Time (t) = 10 seconds

To Find :

• (a) Acceleration

• (b) Distance travelled.

Solution :

By using first equation motion :]

➥ v = u + at

Putting all the given values we get :]

➥ 11.11 = 5.55 × a × 10

➥ 11.11 - 5.55 = 10a

➥ 5.55 = 10a

➥ a = 5.55/10

➥ a = 0.55 m/s²

Therefore,Acceleration produced is 0.55 m/s².

Now, we will find the Distance travelled by car.

• By using third equation of motion :]

➦ v² - u² = 2as

Putting all the given values we get :]

➦ (11.11)² - (5.55)² = 2 × 0.55 × s

➦ 92.62 = 1.1s

➦ s = 84.2 m

Therefore,the Distance travelled by the car is 84.2 m.