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Question
question 12
a runner maintains constant acceleration after starting from rest as she runs a distance of 60.0 m. the runners speed at the end of the 60.0 m is 9.00 m/s. how much time did it take the runner to complete the 60.0 m distance?
9.80 s
13.3 s
10.2 s
15.0 s
Step1: Recall kinematic formula
For constant acceleration, the average velocity formula is \( v_{avg} = \frac{v_0 + v}{2} \), where \( v_0 \) is initial velocity, \( v \) is final velocity. Also, distance \( d = v_{avg} \times t \).
Given \( v_0 = 0 \, \text{m/s} \) (starts from rest), \( d = 60.0 \, \text{m} \), \( v = 9.00 \, \text{m/s} \).
Step2: Calculate average velocity
\( v_{avg} = \frac{0 + 9.00}{2} = 4.50 \, \text{m/s} \)
Step3: Solve for time \( t \)
Using \( d = v_{avg} \times t \), rearrange to \( t = \frac{d}{v_{avg}} \).
Substitute \( d = 60.0 \, \text{m} \), \( v_{avg} = 4.50 \, \text{m/s} \):
\( t = \frac{60.0}{4.50} = 13.3 \, \text{s} \)
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13.3 s (corresponding to the option "13.3 s")