Question

22. A car accelerates at a constant rate from 18 kmh
to 72 km h 'in 10 s. What is the acceleration? If
the car now takes 15 s to stop, what is the
retardation?​

Answer 1

Answer :

• Acceleration = 1.5 m/s²
• Retardation = 1.33 m/s²

Explanation :

Here, we are given that a car accelerates at a constant rate from 18 km/h to 72 km/h in 10 s. So, 18 km/h is its initial velocity and 72 km/h is its final velocity. We have to find the acceleration in the first part.

According to the question,

• Intial velocity (u) is 18 km/h
• Final velocity (v) is 72 km/h.
• Time taken (t) is 10 seconds.

We have to calculate acceleration of the car. For that, we'll be using formula of acceleration. Before that, we need to convert given velocities into its standard form (m/s).

Conversion of units :

• To convert km/h to m/s, we multiply the given value with $\sf \dfrac{5}{18}$

★ Converting initial velocity (u) into its standard form :

$\longrightarrow$ u = 18 × $\sf \dfrac{5}{18}$ m/s

$\longrightarrow$ u = 1 × 5 m/s

$\longrightarrow$ u = 5 m/s

★ Converting final velocity (v) into its standard form :

$\longrightarrow$ v = 72 × $\sf \dfrac{5}{18}$ m/s

$\longrightarrow$ v = 4 × 5 m/s

$\longrightarrow$ v = 20 m/s

We know that, acceleration is the rate of change in velocity with time, here change in velocity is the difference of final velocity and initial velocity.

$\bigstar \: \boxed{\sf {a = \dfrac{v-u}{t} }}$

$\longrightarrow \sf {a = \dfrac{20-5}{10} \: m/s^2}$

$\longrightarrow \sf {a = \dfrac{15}{10} \: m/s^2}$

$\longrightarrow \sf {a = \dfrac{3}{2} \: m/s^2}$

$\longrightarrow \boxed{ \sf {a = 1.5 \: m/s^2}}$

Therefore, acceleration of the car is 1.5 m/s².

Now, in the second part of the question it is stated that the car has same initial velocity as earlier but now it has taken 15 seconds to stop. So, its final velocity will be 0. We've to find its retardation.

Retardation is acceleration with the negative sign. If acceleration will be positive then retardation will be negative and if acceleration will be negative then retardation will be positive.

Finding out the acceleration first. Given,

• Intial velocity (u) is 18 km/h (5m/s)
• Final velocity (v) is 0 m/s [as it stops].
• Time taken (t) is 15 seconds.

$\bigstar \: \boxed{\sf {a = \dfrac{v-u}{t} }}$

$\longrightarrow \sf {a = \dfrac{0-20}{15} \: m/s^2}$

$\longrightarrow \sf {a = \dfrac{-20}{15} \: m/s^2}$

$\longrightarrow \boxed{\sf {a =-1.33 \: m/s^2}}$

So, the acceleration of the car is -1.33 m/s². Therefore, retardation is 1.33 m/s².

$\longrightarrow \boxed{ \sf {Retardation = 1.33 \: m/s^2}}$