Question

VII. Answer the following:-
1. A Car accelerates uniformly from 18kmh in 5s. Calculate
a) The acceleration and b) The distance covered by the car in that time.​

Answer 1

Given :

• Initial velocity (u) = 0 m/s
• Time (t) = 5 seconds
• Final velocity (v) = 18 km/h = 5 m/s

To find :

• Acceleration (a)
• Distance covered (s)

According to the question,

By using Newtons first equation of motion,

⇒ v = u + at

Where,

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

⇒ Substituting the values,

⇒ 5 = 0 + a × 5

⇒ 5 - 0 = 5a

⇒ 5 = 5a

⇒ 5 ÷ 5 = a

⇒ 1 = a

So,the acceleration is 1 m/s².

Now,

⇒ v² = u² + 2as

Where,

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• s = Distance

⇒ (5)² = (0)² + 2 × 1 × s

⇒ 25 = 0 + 2s

⇒ 25 - 0 = 2s

⇒ 25 = 2s

⇒ 25 ÷ 2 = s

⇒ 12.5 = s

So,the distance is 12.5 m.

__________________________

Answer 2

Answer:

Given :

• Initial velocity (u) = 0 m/s

• Final velocity (v) = 18 km/h

• Time (t) = 5 sec

To Find :

1. Acceleration of the car.
2. The distance covered by the car in that time.

Solution :

First convert km/h into m/s.

$: \implies \: \: \: \: \sf \: Final \: velocity = \frac{18 \times 5}{18} \\ \\ \\ : \implies \: \: \: \: \sf \: Final \: velocity = \: 5 \: m s$

$: \implies \: \: \: \boxed{ \green{ \sf \: v = u + at}}$

Substitute all values :

$: \implies \: \: \: \sf \: 5 = 0 + a \times 5 \\ \\ \\ : \implies \: \: \: \sf \: \: 5a = 5 \\ \\ \\ : \implies \: \: \: \sf \:a = \cancel{ \frac{5}{5} } \\ \\ \\ : \implies \: \: \: \sf \: \: a = 1 \: {ms}^{2}$

$: \implies \: \: \: \boxed{ \pink{ \sf \: s= ut + \frac{1}{2} a {t}^{2} }}$

Substitute all values :

$: \implies \: \: \: \sf \: s= 0 \times 5 + \frac{1}{2} \times 1 \times {5}^{2} \\ \\ \\ : \implies \: \: \: \sf \: s= \frac{1}{ \cancel{2} } \times \cancel{ 25 }\\ \\ \\ : \implies \: \: \: \sf \: s= \: 12.5 \: m$

Hence,Solved !!