Question

A body moves in a straight line along the Y-axis. Its distance y (in metres) from the origin is given by \[y = 8t - 3{t^2}.\]The average speed in the time interval from t = 0 second to t = 1 second is
A. \[ - 4\,m/s\]
B. zero
C. \[5\,m/s\]
D. \[6\,m/s\]

Answer 1

Hint:The speed is defined as the rate of change of distance in any direction and it is the magnitude of velocity and it cannot be negative. The distance is the length of the actual path covered by the moving body and hence it cannot be negative. In this case, we need to find the average speed of the moving object.

Formula used :
The average speed of the moving body is,
\[\text{average speed} = \dfrac{\text{total path length}}{\text{total time taken}}\]
Or it can be written as,
\[\text{average speed}({v_{av}}) = \dfrac{{{y_2} - {y_1}}}{{{t_2} - {t_1}}}\]
Where, \[{y_1},{y_2}\] - the distance travelled and \[{t_1},{t_2}\] - time taken.

Complete step by step solution:
The object in movement changes its position with time continuously, so in order to find the change in position with respect to time, we define a quantity called average velocity. But it only relates the displacement of the moving object. So to find the rate of motion along the actual path completely we define average speed.

The average speed describes the rate of motion over the actual path and hence it is defined as the total path travelled divided by the total time taken for the travel.
\[\text{average speed} = \dfrac{\text{total path length}}{\text{total time taken}}\]
The SI unit of average speed is \[m{s^{ - 1}}\] , but it will not give us the direction of the movement, hence it will always be positive.

Given,
\[y = 8t - 3{t^2}\]
To find, \[{v_{av}} = ?\]
The average speed can be given as,
\[\text{average speed}({v_{av}}) = \dfrac{{{y_2} - {y_1}}}{{{t_2} - {t_1}}}\]
\[{v_{av}} = \dfrac{{{y_2} - {y_1}}}{{{t_2} - {t_1}}}\]
So,
\[{y_2} = 8(1) - 3{(1)^2}{\rm{}}\left( {{t_2} = 1s} \right)\]
\[\Rightarrow {y_2} = 8 - 3\]
\[\Rightarrow {y_2} = 5m\]
\[\Rightarrow {y_1} = 8(0) - 3{(0)^2}{\rm{ }}\left( {{t_1} = 0s} \right)\]
\[\Rightarrow {y_1} = 0m\]
\[\Rightarrow {v_{av}} = \dfrac{{5 - 0}}{{1 - 0}}\]
\[\therefore {v_{av}} = 5\,m{s^{ - 1}}\]

So, the correct answer is option C.

Note : The difference between the average velocity and average speed is that , the average speed does not give the direction of the movement and hence it is always positive. But the average velocity can be both positive and negative since it depends on the direction of the displacement. The average speed also has the same SI unit as average velocity\[m{s^{ - 1}}\].