Question

The speed of a car is reduced from 90 km/h to 36 km/h in 5 s. What is the distance travelled by the car during this time interval.

Answer 1

Hint: Since the car is decelerating, the in the opposite direction as its velocity. All units used be converted to SI. The second equation of motion can be used to calculate the distance after calculating the deceleration.

Formula used: In this solution we will be using the following formulae;
\[v = u - at\] where \[a\] is the acceleration of an object, \[v\] is the final velocity, \[u\] is the initial velocity, and \[t\] is the time taken to move from the initial to the final velocity.
\[s = ut - \dfrac{1}{2}a{t^2}\] where \[s\] is the distance covered during an accelerating motion, \[u\] and \[a\] remains initial velocity and acceleration, and \[t\] is the time taken to cover that distance.
\[{v^2} = {u^2} - 2as\], all variables remain as stated prior.

Complete Step-by-Step Solution:
To calculate the acceleration (actually deceleration since the car reduced speed), we shall recall that
\[v = u - at\] where \[a\] is the acceleration of an object, \[v\] is the final velocity, \[u\] is the initial velocity, and \[t\] is the time taken to move from the initial to the final velocity.
Hence, by inserting values, we get
\[10 = 25 - a\left( 5 \right)\] (because 1m/s is equal to \[3.6\] km/h)
Hence, by making \[a\] subject of formula, we have
\[a = \dfrac{{10 - 25}}{{ - 5}} = - \dfrac{{15}}{{ - 5}} = 3m/{s^2}\]
We shall use the equation of motion
\[s = ut - \dfrac{1}{2}a{t^2}\] where \[s\] is the distance covered during an accelerating motion, \[u\] and \[a\] remains initial velocity and acceleration, and \[t\] is the time taken to cover that distance
Hence, by inserting values, we get
\[s = 25\left( 5 \right) - \dfrac{1}{2}\left( 3 \right){5^2}\]
\[ \Rightarrow s = 125 - 37.5 = 87.5m\]

Note: Alternatively, to calculate the distance, one could use the equation of motion
\[{v^2} = {u^2} - 2as\]
Hence, by inserting all known values and making \[s\] subject of the formula, we get
\[{10^2} = {25^2} - 2\left( 3 \right)s\]
\[ \Rightarrow 100 = 625 - 6s\]
Hence,
\[s = \dfrac{{100 - 625}}{{ - 6}} = \dfrac{{525}}{6} = 87.5m\]
Also, note that the sign before the acceleration carries negative signs because the car is decelerating.