WORKED EXAMPLES
Ö Example-1:If a sports car at rest accelerates uniformly to a speed of 144 km h-1 in 5 s then find distance travelled by it ?
Sol : \(
\begin{gathered}
u = 0,v = \,144km\,\,h^{ - 1} \,\,\, = \,\,144\, \times \frac{5}
{{18}}m\,\,s^{ - 1} \, = \,40m\,s^{ - 1} ,t = 5\,\,s \hfill \\
a = \frac{{v - u}}
{t} = \frac{{40}}
{5} = 8m\,\,s^{ - 2} ,\,\,\,\,s = \frac{1}
{2} \times 8 \times (5)^2 = 100\,m \hfill \\
\end{gathered}
\)
Ö Example-2:The driver of a car moving with a velocity of 54 km h-1 applies brakes to decrease its velocity to 36 km h-1 .If the retardation produced by the brakes is 2m s-2, arange the following steps in a sequential order to calculate the distance travelled by the car.
Sol: \(
\begin{gathered}
\,u\, = 54\,km\,\,h^{ - 1} \, = \,54 \times \frac{5}
{{18}} = 15m\,\,s^{ - 1} ,v = 36km\,\,h^{ - 1} \, = 36 \times \frac{5}
{{18}} = 10\,m\,s^{ - 1} ,a = - 2.0\,m\,\,s^{ - 1} \hfill \\
U\sin g\,v^2 \, - \,u^2 = 2as(a) \Rightarrow s = \frac{{v^2 - u^2 }}
{{ - 2a}}(c) \Rightarrow s = \frac{{100 - 225}}
{{ - 2a}}(c) \hfill \\
\Rightarrow s = 125/4 = 31.25m\,(d)\, \hfill \\
\end{gathered}
\)
Ö Example-3:A bike starting from rest picks up a velocity of 72 km h-1 over a distance of 40m.Calculate its acceleration.
Sol : Given, u = 0, v = 72 km h-1 \(
= 72 \times \frac{5}
{{18}} = 20ms^{ - 1}
\) , s = 40m
using v2 - u2 = 2as(20)2 - 0 = 2a x 40 \(
\Rightarrow a = \frac{{400}}
{{2 \times 40}} = 5\,m\,s^{ - 2}
\)
Ö Example-4:A car moving along a straight road with a speed of 72 km h-1 is brought to rest within 3 s after the application of brakes. Calculate the deceleration produced by the brakes.
Sol: Initial velocity ‘u’ =72 km h-1 = \(
72 \times \frac{5}
{{18}} = 20\,\,ms^{ - 1}
\)
Final velocity , v=0 m s-1,
\(
\frac{{v - u}}
{t} = a
\) \(
\Rightarrow \frac{{0 - 20}}
{3} = a \Rightarrow deceleration = 6.67ms^{ - 2}
\)