Question

A car starting from rest acquires a velocity of 54km/h in 5 seconds. Calculate: (i) Its acceleration and (ii) the distance moved by it. ​

Answer 1

Given :-

• A car starting from rest acquires a velocity of 54km/h in 5 seconds.

To find :-

• Acceleration
• Distance moved by car

Solution :-

• Initial velocity (u) = 0

• Final velocity (v) = 54km/h

• Time taken (t) = 5s

→ Final velocity = 54km/h

→ v = 54 × 5/18

→ v = 3 × 5

→v = 15m/s

According first equation of motion

→ v = u + at

→ 15 = 0 + a × 5

→ 15 = 5a

→ a = 15/5

→ a = 3 m/s²

Now, according to second equation of motion

→ s = ut + ½ at²

→ s = 0 × 5 + ½ × 3 × (5)²

→ s = 0 + ½ × 3 × 25

→ s = 75/2

→ s = 37.5m

Hence,

• Acceleration of car is 3m/s²

• Distance coved by car is 37.5m

Answer 2

Answer:

• Acceleration of car = 3 m/s²
• The distance covered by car = 37.5 m

Explanation:

Given that,

• Initial velocity (u) = 0 m/s
• Final velocity (v) = 54 km/h ⇒54 × 5/18 ⇒15 m/s.
• Time (t) = 5 sec

Applying acceleration formula,

⇒ a = (v - u)/t

⇒ a = (15 - 0)/5

⇒ a = 15/5

⇒ a = 3 m/s²

Applying 2nd eqn. of motion,

⇒s = ut + 1/2 × at²

⇒s = 0 × 5 + 1/2 × 3 × (5)²

⇒s = 1/2 × 3 × 25

⇒s = 1.5 × 25

⇒ s = 37.5 m