Question

The velocity or a car changes from 36 kmhr^-1 to 90 kmhr^-1 in 2 Hr. Calculate
a. Change in velocity in ms^-1
b. Acceleration in (i) kmhr^-2​

Answer 1

Answer:

Change in velocity = 15 m/s

Acceleration = 27 km/h²

Explanation:

As per the provided information in the given question, we have :

• Initial velocity (u) = 36 km/h
• Final velocity (v) = 90 km/h
• Time taken (t) = 2 hours

We are asked to calculate change in velocity in m/s and acceleration in km/h².

Calculating change in velocity :

⇒ ∆v = v - u

• v denotes final velocity
• u denotes initial velocity

⇒ ∆v = 90 km/ h - 36 km/h

⇒ ∆v = 54 km/h

Now, let's convert 54 km/h into m/s to find the change in velocity in m/s.

⇒ 1 km/h = $\sf \dfrac{5}{18}$ m/s

⇒ 54 km/h = (54 × $\sf \dfrac{5}{18}$) m/s

⇒ 54 km/h = (3 × 5) m/s

⇒ 54 km/h = 15 m/s

∴ Change in velocity is 15 m/s.

Calculating acceleration in km/h² :

By using the first equation of motion,

$\implies \sf a = \dfrac{v - u}{t}$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time

$\implies \sf a = \dfrac{90 \; kmh^{-1} - 36 ; kmh^{-1}}{2 \; h}$

$\implies \sf a = \dfrac{54\; kmh^{-1} }{2 \; h}$

$\implies \bf a = 27 \; kmh^{-2}$

∴ Acceleration in km/h² is 27 km/h².