Question

10.A car accelerates uniformly from 36 km/h to 72 km/h in 2 seconds. Calculate : a. the acceleration of the car b. the distance covered by the car in that time

Answer 1

$\bigstar$ Answer $\bigstar$

$\checkmark$ Given:

A car accelerates uniformly from 36 km/h to 72 km/h in 2 seconds

$\checkmark$ To Find:

a. Acceleration of the car

b. Distance covered by the car in that time

$\checkmark$ Solution:

We know that,

Acceleration is defined as the rate of change of velocity per unit of time

Mathematically,

$\red{\boxed{\sf a=\left(\dfrac{v-u}{t}\right) \ m/s^{2}}}$

Here,

• a = Acceleration
• u = Initial Velocity
• v = Final Velocity
• t = Time

Given that,

A car accelerates uniformly from 36 km/h to 72 km/h in 2 seconds

Therefore,

Initial Velocity (u) = 36 km/h

Final Velocity (v) = 72 km/h

Converting km/h to m/s

[ divide the speed value by 3.6 ]

We get,

Initial Velocity (u) = 10 m/s

Final Velocity (v) = 20 m/s

Also,

Time (t) = 2 seconds

Hence,

Substituting the values,

We get,

$\blue{\implies \sf a=\dfrac{20-10}{2} \ m/s^{2}}$

$\green{\implies \sf a=\dfrac{10}{2} \ m/s^{2}}$

$\red{\implies \sf a=5 \ m/s^{2}}$

Hence,

a. Acceleration (a) = 5 m/s²

According to the Question,

We are also asked to find the distance covered by the car in that time

Hence, Using

$\star$ Equation of Motion

$\orange{\implies \boxed{\sf s=ut+\dfrac{1}{2}at^{2}}}$

Here,

• s = Distance
• u = Initial Velocity
• t = Time
• a = Acceleration

Given that,

u = 10 m/s

t = 2 2 seconds

And, We found out,

a = 5 m/s²

Hence,

Substituting the values,

We get,

$\blue{\sf \implies s=10(2)+\dfrac{1}{2}\times (5)\times (2)^{2} \ \ m}$

$\green{\sf \implies s=20+\dfrac{1}{2} \times 5 \times 4 \ \ m}$

$\red{\sf \implies s=20+\dfrac{1}{2} \times 20 \ \ m}$

$\orange{\sf \implies s=20+10 \ \ m}$

$\pink{\sf \implies s=30 \ \ m}$

Therefore,

Distance covered by the car in that time is 30 metre

Hence,

b. Distance (s) =  30 m

Answer 2

Answer:

Dist. travelled by a car is 30 m .

Explanation:

u = 36 km/hr = 36 × 5/18 = 10 m/s

v = 72 km/hr = 72 × 5/18 = 20 m/s

t = 2 sec

Now,

a = ( v - u )/t m/s^2

= ( 20 - 10 ) 2 m/s^2

= 10/2 m/s^2

= 5 m/s^2

And

s = ut + 1/2at^2

= 10 × 2 + 1/2 × ( 5 × 2^2 )

= 20 + 1/2 × 20

= 20 + 10

= 30 m

Hence,

The dist. travelled by a car is 30 m.