Question

A jogger runs in a straight line, with a magnitude of average velocity of $5.00m{s}^{-1}$ for $4.00$ minute and then with a magnitude of average velocity of $4.00m{s}^{-1}$ for $3.00$ minute. What is the magnitude of the final displacement from her initial position ?

• A. $1.92\times {10}^{3}m$
• B. $2.56\times {10}^{3}m$
• C. $3.19\times {10}^{3}m$
• D. $0m$

Answer 1

The correct option is A $1.92\times {10}^{3}m$

Answer is A.
Let us imagine that a jogger is running on a track. Notice that the runner runs more slowly on the average during the second time interval as if she is tiring.
Let us use the two separate portions of motion and find out the displacement for each portion.
Velocity = rate of change of displacement/ rate of change of time.
So, Rate of change of displacement = Velocity $times$ rate of change of time.
Therefore, the rate of change of displacement from position 1

= $5.00m/s\times 4.00min\times \frac{60s}{1min}=1.20\times {10}^{3}m$.
The rate of change of displacement from position 2 = $4.00m/s\times 3.00min\times \frac{60s}{1min}=.720\times {10}^{3}m$.
We add up both these displacements to find the total displacement of $1.92\times {10}^{3}m$.
Hence, the magnitude of the final displacement from her initial position is $1.92\times {10}^{3}m$.