What is the average velocity of a particle when it returns to the starting point? Can its average speed be zero?
(N/A) Average velocity is defined as the total displacement divided by the total time taken. When a particle returns to its starting point,the total displacement is $0$. Therefore,the average velocity is $0/t = 0$.
Average speed is defined as the total distance traveled divided by the total time taken. Since distance is a scalar quantity and is always positive for a moving object,the total distance covered cannot be $0$ if the particle has moved. Thus,the average speed cannot be zero.
Average speed is defined as the total distance traveled divided by the total time taken. Since distance is a scalar quantity and is always positive for a moving object,the total distance covered cannot be $0$ if the particle has moved. Thus,the average speed cannot be zero.
Explore More
Similar Questions
$A$ cyclist travels a distance of $4\, km$ from $P$ to $Q$ and then moves a distance of $3\, km$ at a right angle to $PQ$. Find his resultant displacement graphically.
Easy
View Solution$A$ body thrown in the vertically upward direction rises up to a height $h$ and comes back to the position of its start. Calculate: $(a)$ the total distance travelled by the body and $(b)$ the displacement of the body. Under what condition will the magnitude of the displacement be equal to the distance travelled by an object?
Medium
View SolutionThe graph given below is the distance-time graph of an object.
$(i)$ Find the speed of the object during the first four seconds of its journey.
(ii) How long was it stationary?
(iii) Does it represent a real-life situation? Justify your answer.
$(i)$ Find the speed of the object during the first four seconds of its journey.
(ii) How long was it stationary?
(iii) Does it represent a real-life situation? Justify your answer.
Difficult
View SolutionAn object is moving along a straight line with uniform acceleration. The following table gives the velocity of the object at various instants of time.
Time $(s)$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ Velocity $(m s^{-1})$ $2$ $4$ $6$ $8$ $10$ $12$ $14$
Plot the graph.
From the graph:
$(i)$ Find the velocity of the object at the end of $2.5 \ s$.
$(ii)$ Calculate the acceleration.
$(iii)$ Calculate the distance covered in the last $4 \ s$.
| Time $(s)$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ |
| Velocity $(m s^{-1})$ | $2$ | $4$ | $6$ | $8$ | $10$ | $12$ | $14$ |
Plot the graph.
From the graph:
$(i)$ Find the velocity of the object at the end of $2.5 \ s$.
$(ii)$ Calculate the acceleration.
$(iii)$ Calculate the distance covered in the last $4 \ s$.
Medium
View SolutionThe driver of a train $A$ travelling at a speed of $54\, km\, h^{-1}$ applies brakes and retards the train uniformly. The train stops in $5\, s$. Another train $B$ is travelling on the parallel track with a speed of $36\, km\, h^{-1}$. This driver also applies the brakes and the train retards uniformly. The train $B$ stops in $10\, s$. Plot the speed-time graph for both the trains on the same paper. Also,calculate the distance travelled by each train after the brakes were applied.
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo