Question

A car accelerated uniformly from 18 km/h to 36 km/h in 5 sec. Calculate the distance covered by the car in that time

Answer 1

Answer:

37.5 m

Explanation:

According to the given question,

u = initial velocity = 18 km/h = 18×1000/3600 = 5 m/s

v = final velocity = 36 km/h = 36×1000/3600 = 10 m/s

t = time = 5 sec

s = displacement

But here displacement can be treated as distance as it's linear motion.

Thus, s = distance / displacement

We know,

• s = (u+v)/2 × t

(This is the 5th formula of accelerated motion)

Putting the values we get,

• s = (5+10)/2 × 5

• s = 15/2 × 5

• s = 75/2

• s = 37.5 m

Therefore, distance travelled by the car is 37.5 m from the given data.

Alternatively

We know,

• v = u + at

Putting the values we get,

• 10 = 5 + 5a

• 5a = 10-5

• 5a = 5

• a = 1 m/s^2

Again,

• s = ut + 1/2 at^2

Putting the values we get,

• s = 5×5 + 1/2 × 1 × 25

• s = 25 + 25/2

• s = (50+25)/2

• s = 75/2

• s = 37.5 m

Which is same as the 5th formula of motion.

(5th formula of motion is the only formula where there is no acceleration term in it, so it's very important. And this formula is written in very few books.)

Answer 2

Solution:

Given:

A car accelerated uniformly from 18 km/h to 36 km/h in 5 sec.

Find:

Calculate the distance covered by the car in that time.

Calculations:

1. u = initial velocity.
2. v = final velocity.
3. t = time.
4. s = displacement.

Formula (1)

→ s = distance/displacement.

Formula (2)

→ s = (u + v)/2 × t

Adding values to equation:

→ s = (5 + 10)/2 × 5

→ s = 15/2 × 5

→ s = 75/2

→ s = 37.5

Hence, 37.5 m is the distance travelled by the car.

Alternatively :

→ v = u + at

Adding values to equation:

→ 10 = 5 + 5a

→b5a = 10 - 5

→b5a = 5

→ a = 1 m/s²

→ s = ut + 1/2 at²

Adding values to equation:

→ s = 5 × 5 + 1/2 × 1 × 25

→ s = 25 + 25/2

→ s = (50 + 25)/2

→ s = 75/2

→ s = 37.5 m

Hence, 37.5 m is the distance travelled by the car.

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