Question

A car accelerates uniformly from 30 km/h to 60 km/h in 5 seconds Calculate (i) acceleration in m/s2 and (ii) distance covered by the car in metres
during this interval.​

Answer 1

Answer:

(i) Acceleration is 1.67 m/s² .

(ii) Distance covered by car is 62.525 m.

Explanation:

To find :

1. Acceleration in m/s².
2. Distance covered by car.

Given that,

• Initial speed, u is 30 km/h
• Final speed, v is 60 km/h.
• Time period, t is 5 seconds.

(i) We know,

• Acceleration = v - u/t .

Conversion :

• 30 km/h = (30 × 5)/18 = 8.34 m/s

• 60 km/h = (60 × 5)/18 = 16.67 m/s .

Now, Put values in formula :

→ Acceleration = 16.67 - 8.34/5

→ Acceleration = 8.33/5

→ Acceleration= 1.67

Thus,

Acceleration is 1.67 m/s² .

(ii) We know,

Second equation of motion :

• s = ut + 1/2 at²

[Where, s is distance/displacement]

Put all values in this equation :

→ s = (8.34 × 5) + 1/2 × (1.67) × (5)²

→ s = 41.7 + 1/2 × 1.67 × 25

→ s = 41.7 + 41.75/2

→ s = (83.4 + 41.75)/2

→ s = 125.15/2

→ s = 62.525

Thus,

Distance covered by car is 62.525 m.

Answer 2

Question:

• A car accelerates uniformly from 30 km/h to 60 km/h in 5 seconds Calculate ::

• (i) Acceleration in m/s²?
• (ii) Distance covered by the car in metres during this interval?

Answer:

• (i) Acceleration of car in m/s² is 1.66 m/s².
• (ii) Distance covered by the car in metres during this interval is 62.4 meters.

Explanation:

Given that:

• Initial velocity (u) = 30 km/h = (30 × 5/18) m/s = 8.33 m/s
• Final velocity (v) = 60 km/h = (60 × 5/18) m/s = 16.66 m/s
• Tine taken (t) = 5 seconds

To Find:

• Acceleration (a)?
• Distance covered (s)?

Solution:

• Using first equation of motion ::

We know that,

⋆ $\boxed{\bf{v = u + at}}$

According to the question by using the formula we get,

➻ $\sf 16.66 = 8.33 + (a\:\times\:5)$

➻ $\sf 16.66 - 8.33 = 5a$

➻ $\sf 5a = 8.33$

➻ $\sf a = {\cancel{\dfrac{8.33}{5}}}$

➻ $\bf\red{a = 1.66\:m/s^2}$

∴ Hence, acceleration of car in m/s² is 1.66 m/s².

• Now, let's find distance covered by the car. Using second equation of motion ::

We know that,

⋆ $\boxed{\bf{s = ut + \dfrac{1}{2}at^2}}$

According to the question by using the formula we get,

➻ $\sf s = (8.33\:\times\:5) + \dfrac{1}{\cancel{2}}\:\times\:\cancel{1.66}\:\times\:(5)^2$

➻ $\sf s = 41.65 + (0.83\:\times\:35)$

➻ $\sf s = 41.65 + 20.75$

➻ $\bf\purple{s = 62.4\:m}$

∴ Distance covered by the car in metres during this interval is 62.4 meters.

Know More:

$\clubsuit$ Three equations of motion ::

• v = u + atㅤㅤ ㅤ ㅤ [Used above]
• s = ut + ½ at²ㅤㅤㅤ[Used above]
• v² = u² + 2as

$\clubsuit$ Some important definitions ::

• Acceleration

Acceleration is the process where velocity changes. Since, velocity is the speed and it has some direction. So, change in velocity is considered as acceleration.

• Initial velocity

Initial velocity is the velocity of the object before the effect of acceleration.

• Final velocity

After the effect of acceleration, velocity of the object changes. The new velocity gained by the object is known as final velocity.

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