Question

-1
Example 8.6 A car accelerates uniformly
from 18 km h to 36 km h in 5 s.
Calculate (i) the acceleration and (ii) the
distance covered by the car in that time.​

Answer 1

Given:-

• Initial Velocity ,u = 18km/h = 18×5/18 = 5m/s

• Final velocity ,v = 36km/h = 36×5/18 = 10 m/s

• Time taken ,t = 5 s

To Find:-

• The acceleration ,a
• Distance covered by the car in that time ,s

Solution:-

As we know that Acceleration is defined as the rate of change of velocity at per unit time.

• a = v-u/t

v is the final velocity

a is the acceleration

u is the initial velocity

t is the time taken

Substitute the value we get

→ a = 10-5/5

→ a = 5/5

→ a = 1m/s²

Therefore,the acceleration of the car is 1 m/s².

Now, we have to calculate the distance covered by the car.

Using 3rd Equation of Motion

• v² = u² +2as

substitute the value we get

→ 10² = 5² + 2×1 ×s

→ 100 = 25 + 2 × s

→ 100-25 = 2×s

→ 75 = 2×s

→ s = 75/2

→ s = 37.5 m

Therefore,the distance covered by the car is 37.5 metres.

Answer 2

Answer:

Given :-

• A car accelerates uniformly from 18 km/h to 36 km/h in 5 seconds.

To Find :-

• Acceleration.
• Distance covered by the car in that time.

Formula Used :-

❶ To find acceleration,

✦ v = u + at

❷ To find distance covered,

❖ s = ut + ½ at² ❖

Solution :-

Given :-

» Initial velocity = 18 km/h = 18 × 5/18 m/s = 5 m/s

» Final velocity = 36 km/h = 36 × 5/18 m/s = 10 m/s

» Time = 5 secs

❶ To find acceleration

⇒ v = u + at

⇒ 10 = 5 + a(5)

⇒ 10 = 5 + 5a

⇒ 10 = 10a

⇒ 10 ÷ 10 = a

➠ 1 = a

∴ The acceleration of the car is 1 m/s² .

❷ To find distance covered by the car,

↦ s = ut + ½ at²

↦ s = (5)(5) + ½ × (1)(5)²

↦ s = 25 + ½ × 25

↦ s = 25 + 12.5

➥ s = 37.5 m

∴ The distance covered by the car in that time is 37.5 m .