Q2.8-65PE
Question
A graph of v(t) is shown for a world-class track sprinter in a \({\bf{100}} - {\bf{m}}\) race. (See Figure 2.67). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous velocity at \(t = {\rm{ }}{\bf{5}}{\rm{ }}{\bf{s}}\)? (c) What is his average acceleration between 0 and 4 s? (d) What is his time for the race?
Step-by-Step Solution
Verified(a) \(6 m/s\)
(b) \(12{\rm{ m/s}}\)
(c) The acceleration is approximately to the value \(3 m/{s^2}\)
(d) \(10.3 s\)
To analyses movement, motion graphs can be employed.
The mathematical approaches and graphical solutions for determining motion equations are the same.
The slope of a graph of displacement x vs. time t represents velocity v.
The inclination of a graph of velocity v vs. time t is called acceleration.
Average velocity, instantaneous velocity, and acceleration can all be calculated using graphs.
a) Here if we carefully look into the figure, in your text book lets take any two points on the line
\(\begin{array}{c}average\,velocity = \frac{{{v_1} + {v_2}}}{2}\\ = \frac{{0 + 12}}{2}\\ = 6\,m/s\end{array}\)
Hence the average velocity of the body up to 4 s is 6 m/s
b)
Hence when the time is 5s the velocity is constantly 12 m/s.
(c)
Here if we carefully look into the figure, in your text book let’s take any two points on the line
Coordinate \(\left( {0,0} \right)\) and \(\left( {4,12} \right)\)
Slop of the line will be
\(\begin{array}{l}m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_2}}}\\Here\,the\,coordinates\,are\,(0,0)\,\,\,and\,\,\,(4,12)\\m = \frac{{12 - 0}}{{4 - 0}}\\m = 3\,m/{s^2}\end{array}\)
Hence the acceleration is approximately to the value\(3 m/{s^2}\).
(d) For the first \(4{\rm{ s}}\)
Distance travelled is equal to area under the curve
Distance = \(\frac{1}{2} \times 4 \times 12 = 24m\)
He can run the remaining distance by
\(\begin{array}{c}t = \frac{{distance}}{{velocity}} = \frac{{76}}{{12}}\\ = 6.3 s\end{array}\)
Total time of the sprint is\({t_{total}} = 4 + 6.3 = 10.3 s\)