In a long distance race,the athletes were expected to take four rounds of the track such that the finish line was the same as the start line. Suppose the length of the track was $200 \ m$.
$(a)$ What is the total distance to be covered by the athletes?
$(b)$ What is the displacement of the athletes when they touch the finish line?
$(c)$ Is the motion of the athletes uniform or nonuniform?
$(d)$ Is the displacement of an athlete and the distance moved by him at the end of the race equal?
$(a)$ What is the total distance to be covered by the athletes?
$(b)$ What is the displacement of the athletes when they touch the finish line?
$(c)$ Is the motion of the athletes uniform or nonuniform?
$(d)$ Is the displacement of an athlete and the distance moved by him at the end of the race equal?
(D) The total distance is $200 \ m \times 4 = 800 \ m$.
$(b)$ Since the athletes return to the starting point,their displacement is $0 \ m$.
$(c)$ Since the velocity of the athletes changes throughout the race,their motion is nonuniform.
$(d)$ No,the displacement $(0 \ m)$ and the distance travelled $(800 \ m)$ are not equal.
$(b)$ Since the athletes return to the starting point,their displacement is $0 \ m$.
$(c)$ Since the velocity of the athletes changes throughout the race,their motion is nonuniform.
$(d)$ No,the displacement $(0 \ m)$ and the distance travelled $(800 \ m)$ are not equal.
Explore More
Similar Questions
$(a)$ Write differences between speed and velocity.
$(b)$ The speed of a moving object is $x \ m \ s^{-1}$. Its velocity is $y \ m \ s^{-1}$. What is the similarity observed?
$(b)$ The speed of a moving object is $x \ m \ s^{-1}$. Its velocity is $y \ m \ s^{-1}$. What is the similarity observed?
Easy
View Solution$A$ girl of mass $50\, kg$ jumps out of a moving boat of mass $300\, kg$ on to the bank with a horizontal velocity of $3\, m s^{-1}$. With what velocity does the boat begin to move backwards (in $, m s^{-1}$)?
Difficult
View SolutionThe area under the velocity$-$time graph gives the value of
Easy
View SolutionArea under a speed-time graph is equal to the
Easy
View Solution$A$ train starting from rest picks up a speed of $10\, m s^{-1}$ in $100\, s$. It continues to move at the same speed for the next $250\, s$. It is then brought to rest in the next $50\, s$. Plot a speed-time graph for the entire motion of the train and calculate:
$(i)$ acceleration of the train while accelerating,
$(ii)$ retardation of the train while retarding,
$(iii)$ the total distance covered by the train.
$(i)$ acceleration of the train while accelerating,
$(ii)$ retardation of the train while retarding,
$(iii)$ the total distance covered by the train.
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