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Question: Joseph Jogs from one end \(A\) to the other end \(B\) of a straight \(300\,m\) road in \(2\,minutes\...

Joseph Jogs from one end AA to the other end BB of a straight 300m300\,m road in 2minutes30seconds2\,minutes\,30\,seconds and then turns around and jogs 100m100\,m back to point CC in another 1minute1\,minute. What are Joseph’s average speeds and velocities in jogging (a) from AA to BB and (b) from AA to CC?

Explanation

Solution

This is a very simple question of mathematics, as all the information to get the average speed and average velocity is given in the question, just read the question properly, write the data in a sequential way and then getting the answer is easy. The formula of average speed and average velocity should be known, as this is a basic formula that is widely used in physics.

Formula used:
Average Speed=Distance CoveredTime Taken Average Velocity=DisplacementTime Taken \begin{aligned} & \text{Average Speed}=\dfrac{\text{Distance Covered}}{\text{Time Taken}} \\\ & \text{Average Velocity}=\dfrac{\text{Displacement}}{\text{Time Taken}} \\\ \end{aligned}

Complete Step-by-Step solution :
Distance covered from AA to BB = 300m300\,m and since Joseph is jogging in a straight line, so
Displacement=300m\text{Displacement}=300\,m

Time taken to jog from AA to BB = 2minutes30seconds=(2×60)+30=150sec2\,minutes\,30\,seconds\,=(2\times 60)+30=150\,sec
Therefore,
Average Speed=Distance CoveredTime Taken=300150=2m/sec Average Velocity=DisplacementTime Taken=300150=2m/sec \begin{aligned} & \text{Average Speed}=\dfrac{\text{Distance Covered}}{\text{Time Taken}}=\dfrac{300}{150}=2\,m/sec \\\ & \text{Average Velocity}=\dfrac{\text{Displacement}}{\text{Time Taken}}=\dfrac{300}{150}=\,\,2\,m/sec \\\ \end{aligned}

Distance covered from AA to CC = (300+100)m=400m(300\,+\,100)\,m=400\,m and since Joseph is jogging in a straight line, so
Displacement=(300100)m=200m\text{Displacement}=(300\,-100)\,m=200\,m

Time taken to jog from AA to CC = 2minutes30seconds+1minute=3minutes30seconds=(3×60)+30=210sec2\,minutes\,30\,seconds\,+\,1\,minute=3\,minutes\,30\,seconds=(3\times 60)+30=210\,sec

Therefore,
Average Speed=Distance CoveredTime Taken=400210=1.90m/sec Average Velocity=DisplacementTime Taken=200210=0.95m/sec \begin{aligned} & \text{Average Speed}=\dfrac{\text{Distance Covered}}{\text{Time Taken}}=\dfrac{400}{210}=1.90\,m/sec \\\ & \text{Average Velocity}=\dfrac{\text{Displacement}}{\text{Time Taken}}=\dfrac{200}{210}=\,\,0.95\,m/sec \\\ \end{aligned}

Hence, Average Speed and Average Velocity will depend on the distance covered, displacement and the time taken.

Additional Information:
The velocity of an object is the rate of change of position with respect to a frame of reference, and is a function of time. Velocity is equivalent to a specification of an object's speed and direction of motion (e.g. 60 km/h to the north). Velocity is a fundamental concept in kinematics and physics. It is a branch of classical mechanics that describes the motion of bodies.

Note:
The formula of average speed and average velocity, is the most basic formula that is used in physics. This is the most basic question that a student can get in the exam and these questions are bonus questions, as it helps to increase marks in the exam. And there are many advanced topics in physics, which uses this concept as the base to learn those advanced topics.