Question

A car accelerates from 8m/s to 16m/s in 2
minutes. Calculate the acceleration and the
distance covered by the car.

Answer 1

Answer:

2min = 120s

a = v-u/t

= 16-8/120

= 8/120

= 0.64 .

s= ut+ 1/2at²

= 8×2 + 1/2(0.64)×(2)²

= 16+ 1.28

= 17.28 (approx)

Answer 2

Given:-

• Initial velocity ,u = 8m/s
• Final velocity ,v = 16m/s
• Time taken ,t = 2min

To Find:-

• Acceleration ,a
• Distance covered ,s

Solution:-

According to the Question

It is given that car accelerates from 8m/s to 16m/s in 2

minutes.

Firstly we convert the unit here

→ 1min = 60

→ 2min = 2×60 = 120s

Therefore ,t = 120s

Now, calculating the acceleration

As we know that acceleration is defined as the rate of change in velocity.

• a = v-u/t

where,

• 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 = 16-8/120

→ a = 8/120

→ a = 0.066m/s²

• Hence, the acceleration of the car is 0.066m/s²

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Calculating the distance covered by the car .

Using 3rd equation of motion

• v² = u² + 2as

Substitute the value we get

→ 16² = 8² + 2×0.066 × s

→ 256 = 64 + 0.132 s

→ 256-64 = 0.132 × s

→ 192 = 0.132 × s

→ s = 192/0.132

→ s = 1454.54m

• Hence, the distance covered by the car is 1454.54 metres .

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