Question

1. a graph of v(t) is shown for a world - class track sprinter in a 100 - m race. (see figure 2.65). (a) what is his average velocity for the first 4 s? (b) what is his instantaneous velocity at t = 5 s? (c) what is his average acceleration between 0 and 4 s? (d) what is his time for the race?

runner velocity vs. time
graph with velocity (m/s) on y - axis (0, 2, 4, 6, 8, 10, 12, 14) and time (s) on x - axis (0, 2, 4, 6, 8, 10, 12); velocity increases linearly from (0,0) to (4,12) then remains constant

Explanation:

Part (a): Average Velocity for First 4 s

Step1: Recall average velocity formula

Average velocity \( v_{avg} = \frac{\text{displacement}}{\text{time}} \). Displacement from \( v-t \) graph is area under the curve. For \( 0 - 4 \) s, the graph is a triangle with base \( t = 4 \) s and height \( v = 12 \) m/s (from the graph, at \( t = 4 \) s, velocity is 12 m/s? Wait, no, wait: Wait, the graph: from \( t=0 \) to \( t=4 \) s, it's a straight line from (0,0) to (4,12)? Wait, no, looking at the graph, at \( t=2 \) s, velocity is 6 m/s, so slope is \( \frac{6}{2} = 3 \) m/s², so at \( t=4 \) s, velocity is \( 3 \times 4 = 12 \) m/s. So the area under \( v-t \) from 0 to 4 s is area of triangle: \( \frac{1}{2} \times base \times height = \frac{1}{2} \times 4 \times 12 \).

Step2: Calculate displacement

Displacement \( \Delta x = \frac{1}{2} \times 4 \, \text{s} \times 12 \, \text{m/s} = 24 \, \text{m} \).

Step3: Calculate average velocity

Average velocity \( v_{avg} = \frac{\Delta x}{\Delta t} = \frac{24 \, \text{m}}{4 \, \text{s}} = 6 \, \text{m/s} \).

Part (b): Instantaneous Velocity at \( t = 5 \) s

Step1: Analyze the graph

After \( t = 4 \) s, the velocity is constant (horizontal line). So at \( t = 5 \) s, which is after 4 s, the velocity is the same as the constant velocity after 4 s. From the graph, the constant velocity is 12 m/s (since at \( t=4 \) s, velocity is 12 m/s and remains constant).

Part (c): Average Acceleration between 0 and 4 s

Step1: Recall acceleration formula

Average acceleration \( a_{avg} = \frac{\Delta v}{\Delta t} \).

Step2: Find \( \Delta v \) and \( \Delta t \)

Initial velocity \( v_0 = 0 \) m/s (at \( t=0 \)), final velocity \( v_4 = 12 \) m/s (at \( t=4 \) s). Time interval \( \Delta t = 4 - 0 = 4 \) s.

Step3: Calculate acceleration

\( a_{avg} = \frac{12 - 0}{4 - 0} = \frac{12}{4} = 3 \, \text{m/s}^2 \).

Part (d): Time for the Race

Answer:

s:
(a) \(\boldsymbol{6 \, \text{m/s}}\)
(b) \(\boldsymbol{12 \, \text{m/s}}\)
(c) \(\boldsymbol{3 \, \text{m/s}^2}\)
(d) \(\boldsymbol{\frac{31}{3} \, \text{s} \approx 10.33 \, \text{s}}\)