Question

1. how long will it take a car to go from a complete stop to 44 km/hr if they are accelerating at 5 km/hr²?
2. how long will it take a car to accelerate from 15.2 m/s to 23.5 m/s if the car has an average acceleration of 3.2 m/s²?

Explanation:

Step1: Convert units

First, convert \(44\ km/hr\) to \(m/s\).
We know that \(1\ km = 1000\ m\) and \(1\ hr=3600\ s\).
So, \(v = 44\times\frac{1000}{3600}\ m/s=\frac{440}{36}\ m/s=\frac{110}{9}\ m/s\approx12.22\ m/s\)

Step2: Use the formula \(v = v_0+at\) (here \(v_0 = 0\) as it starts from rest)

The formula for acceleration \(a=\frac{v - v_0}{t}\), since \(v_0 = 0\), we can rewrite it as \(t=\frac{v}{a}\)
Given \(a = 5\ km/hr^2\). Convert \(a\) to \(m/s^2\).
\(a=5\times\frac{1000}{3600\times3600}\ m/s^2=\frac{5000}{12960000}\ m/s^2=\frac{5}{12960}\ m/s^2\)
Using \(t=\frac{v}{a}\), substitute \(v=\frac{110}{9}\ m/s\) and \(a = \frac{5}{12960}\ m/s^2\)
\(t=\frac{\frac{110}{9}}{\frac{5}{12960}}=\frac{110}{9}\times\frac{12960}{5}\)
\(t = 110\times\frac{12960}{45}=110\times288 = 31680\ s\)
Convert \(s\) to \(hr\): \(t=\frac{31680}{3600}=8.8\ hr\)

Answer:

It will take \(8.8\ hr\)