Question

A scooterist covers 3km in 5 minutes. Calculate his speed.
$\begin{aligned} & A.\text{ in cm/s} \\\ & B.\text{ in m/s} \\\ & C.\text{ in km/h} \\\ \end{aligned}$

Answer 1

We know that speed of velocity is the rate of change of position of an object with time. Firstly calculate velocity in m/s by dividing distance and time then, find the speed of scooterist in cm/s and km/h.

Formula used:
$S=\dfrac{d}{t}$

Complete step by step solution:
We know that speed gives us the idea of how fast an object is moving with change in time. It can be defined as the distance covered by an object in a given period of time. So, $S=\dfrac{d}{t}$ Where ‘S’ is the velocity of a particle, ‘d’ is the distance covered and ‘t’ is the time. It is given that t = 5min and d = 3km. Now find the speed of the scooter firstly change ‘d’ in terms of m and cm. time in sec and hour.
Given, $d=3km=3000m=3000\times {{10}^{2}}$
$t=5\min =5\times 60=300\sec =\dfrac{1}{12}hrs$
$\begin{aligned} & (a)\text{ }S=\dfrac{d}{t}=\dfrac{3000\times 100}{300} \\\ & S=1000cm/s \\\ \end{aligned}$
The speed of scooterist is 1000cm/s.
$(b)\text{ }S=\dfrac{3000}{300}=10m/s$
The speed of scooterist is 10m/s.
$\begin{aligned} & (c)\text{ }S=\dfrac{3}{\dfrac{1}{12}}=\dfrac{3\times 12}{1} \\\ & S=36km/hr \\\ \end{aligned}$

Hence, the speed of scooterist is 36km/hr.

Additional Information:
Speed is a scalar quantity. This means that speed does not have a direction associated with it. The vector form of speed is known as velocity. Velocity has both magnitude as well as direction associated with it. Speed obeys the basic mathematical algebra. On the other hand, velocity obeys vector algebra only and does not obey the basic mathematical algebra.
The difference between velocity and speed can be clearly understood by the example that in circular motion speed might remain constant, but velocity changes at every point

Note:
It is advised to remember that 1km = 1000m, 1m = 100cm and 1min = 60sec, $1\sec =\dfrac{1}{60\times 60}hr$ that is $1\sec =\dfrac{1}{3600}hr$. These are used while solving this question to calculate the speed we know that
$\text{speed}=\dfrac{\text{distance travelled}}{\text{time taken}}$ .