Question

1. a runner speeds up from 5.00 m/s to 12.0 m/s in a time of 12.0 seconds.

a. calculate the runner’s displacement.

b. calculate the runner’s acceleration.

Explanation:

Part a: Calculate the runner’s displacement

The runner is moving with constant acceleration (since speed is changing uniformly), so we can use the kinematic equation for displacement with constant acceleration: \( s = v_0t+\frac{1}{2}at^2 \), but first we need to find acceleration (or we can also use the average velocity formula \( s=\frac{v_0 + v}{2}t \), where \( v_0 \) is initial velocity, \( v \) is final velocity, and \( t \) is time, since for constant acceleration, average velocity is the average of initial and final velocities).

Step 1: Identify the given values

Initial velocity, \( v_0 = 5.00\space m/s \)
Final velocity, \( v = 12.0\space m/s \)
Time, \( t = 12.0\space s \)

Step 2: Use the average velocity formula for displacement

The average velocity \( \bar{v}=\frac{v_0 + v}{2} \), and displacement \( s=\bar{v}t \).
Substitute the values:
\( \bar{v}=\frac{5.00 + 12.0}{2}=\frac{17.0}{2} = 8.50\space m/s \)
Then, \( s=\bar{v}t=8.50\times12.0 \)

Step 3: Calculate the displacement

\( s = 102\space m \)

Part b: Calculate the runner’s acceleration

Acceleration is the rate of change of velocity, so we use the formula \( a=\frac{v - v_0}{t} \), where \( v \) is final velocity, \( v_0 \) is initial velocity, and \( t \) is time.

Step 1: Identify the given values

Initial velocity, \( v_0 = 5.00\space m/s \)
Final velocity, \( v = 12.0\space m/s \)
Time, \( t = 12.0\space s \)

Step 2: Apply the acceleration formula

\( a=\frac{v - v_0}{t}=\frac{12.0 - 5.00}{12.0} \)

Step 3: Calculate the acceleration

\( a=\frac{7.00}{12.0}\approx0.583\space m/s^2 \)

Final Answers:

a. The runner’s displacement is \(\boldsymbol{102\space m}\) (or more precisely, using the first method with acceleration: first find \( a=\frac{12 - 5}{12}=\frac{7}{12}\space m/s^2 \), then \( s = 5\times12+\frac{1}{2}\times\frac{7}{12}\times12^2=60 + 42 = 102\space m \), same result).
b. The runner’s acceleration is \(\boldsymbol{\approx0.583\space m/s^2}\) (or \(\frac{7}{12}\space m/s^2\approx0.583\space m/s^2\)).

Answer:

Acceleration is the rate of change of velocity, so we use the formula \( a=\frac{v - v_0}{t} \), where \( v \) is final velocity, \( v_0 \) is initial velocity, and \( t \) is time.

Step 1: Identify the given values

Initial velocity, \( v_0 = 5.00\space m/s \)
Final velocity, \( v = 12.0\space m/s \)
Time, \( t = 12.0\space s \)

Step 2: Apply the acceleration formula

\( a=\frac{v - v_0}{t}=\frac{12.0 - 5.00}{12.0} \)

Step 3: Calculate the acceleration

\( a=\frac{7.00}{12.0}\approx0.583\space m/s^2 \)

Final Answers:

a. The runner’s displacement is \(\boldsymbol{102\space m}\) (or more precisely, using the first method with acceleration: first find \( a=\frac{12 - 5}{12}=\frac{7}{12}\space m/s^2 \), then \( s = 5\times12+\frac{1}{2}\times\frac{7}{12}\times12^2=60 + 42 = 102\space m \), same result).
b. The runner’s acceleration is \(\boldsymbol{\approx0.583\space m/s^2}\) (or \(\frac{7}{12}\space m/s^2\approx0.583\space m/s^2\)).