Question

4.Find acceleration ofa car from Start attain velocity of 72km/hint time 4sec.?​

Answer 1

Appropriate Question:

Find acceleration of a car that from start attains velocity of 72km/h in time 4 seconds ?

Answer:

5 m/s²

Explanation:

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

• Initial velocity (u) = 0 m/s (As it starts from rest)
• Final velocity (v) = 72 km/h
• Time taken (t) = 4 seconds

Before commencing the steps, let's first convert final velocity in m/s.

$\longmapsto \rm {1 \; kmh^{-1} = \dfrac{5}{18} \; ms^{-1} }$

$\longmapsto \rm {72\; kmh^{-1} = \Bigg ( 72 \times \dfrac{5}{18} \Bigg ) \; ms^{-1} }$

$\longmapsto \rm {72\; kmh^{-1} = \Big ( 4 \times 5 \Big ) \; ms^{-1} }$

$\longmapsto \bf {72\; kmh^{-1} = 20 \; ms^{-1} }$

$\rule{200}2$

Now, finding the acceleration. Acceleration is defined as the rate of change in velocity. Its SI unit m/s². It is a vector quantity, that means it requires both magnitude and direction for its description.

Method 1 :

By using the formula of acceleration :

$\longmapsto \bf {a = \dfrac{v-u}{t} }$

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

$\longmapsto \rm {a = \Bigg ( \dfrac{20-0}{4} \Bigg ) \; ms^{-2} }$

$\longmapsto \rm {a = \Bigg ( \dfrac{20}{4} \Bigg ) \; ms^{-2} }$

$\longmapsto \bf {a = 5 \; ms^{-2} }$

∴ Acceleration is 5 m/s².

$\rule{200}2$

Method 2 :

By using the first equation of motion :-

$\longmapsto \bf {v= u + at}$

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

$\longmapsto \rm { 20 = 0 + 4a}$

$\longmapsto \rm { 20-0 = 4a}$

$\longmapsto \rm { 20= 4a}$

$\longmapsto \rm { \dfrac{20}{4} = a}$

$\longmapsto \bf { 5= a}$

∴ Acceleration is 5 m/s².