A sprinter can accelerate with constant acceleration for 4.00 s before reaching top speed. He can run the 100 meter dash in 10.00 s .
The correct answer and explanation is:
Let’s analyze the problem step by step:
Problem Recap:
- A sprinter accelerates with constant acceleration for 4.00 seconds.
- After that, the sprinter runs at top speed (constant velocity) for the rest of the race.
- The total time for the 100-meter dash is 10.00 seconds.
We want to find quantities like:
- The sprinter’s acceleration,
- The top speed,
- And the distance covered during acceleration and at top speed.
Step 1: Define variables
- aa = acceleration (constant) during first 4 seconds
- t1=4.00 st_1 = 4.00 \, s = acceleration phase time
- t2=10.00−4.00=6.00 st_2 = 10.00 – 4.00 = 6.00 \, s = constant velocity phase time
- vv = top speed reached after acceleration phase
- d1d_1 = distance covered during acceleration
- d2d_2 = distance covered during constant velocity phase
- Total distance d=100 md = 100 \, m
Step 2: Use kinematic equations for acceleration phase
The sprinter starts from rest, so initial velocity u=0u = 0.
Final velocity after acceleration time t1t_1 is: v=u+at1=a×4v = u + a t_1 = a \times 4
Distance covered during acceleration d1d_1 is: d1=ut1+12at12=12a×42=8ad_1 = u t_1 + \frac{1}{2} a t_1^2 = \frac{1}{2} a \times 4^2 = 8a
Step 3: Use constant velocity phase equations
After acceleration, the sprinter runs at velocity vv for time t2=6t_2 = 6 s, covering distance: d2=v×t2=(4a)×6=24ad_2 = v \times t_2 = (4a) \times 6 = 24a
Step 4: Total distance equation
The total distance is 100 m, so: d1+d2=100d_1 + d_2 = 100 8a+24a=1008a + 24a = 100 32a=10032a = 100 a=10032=3.125 m/s2a = \frac{100}{32} = 3.125 \, m/s^2
Step 5: Calculate top speed
v=a×4=3.125×4=12.5 m/sv = a \times 4 = 3.125 \times 4 = 12.5 \, m/s
Final answers:
- Acceleration a=3.125 m/s2a = 3.125 \, m/s^2
- Top speed v=12.5 m/sv = 12.5 \, m/s
- Distance during acceleration d1=8a=25 md_1 = 8a = 25 \, m
- Distance during constant velocity d2=24a=75 md_2 = 24a = 75 \, m
Explanation in 300 words:
This problem models a sprinter’s motion as two phases: acceleration with a constant rate for 4 seconds, followed by running at a constant top speed for the remaining time. The total distance is 100 meters, and the total time taken is 10 seconds.
First, we apply the kinematic equations to the acceleration phase. Since the sprinter starts from rest, the velocity after accelerating for 4 seconds is v=a×4v = a \times 4. The distance covered during acceleration can be found with the formula d=12at2d = \frac{1}{2} a t^2, which yields d1=8ad_1 = 8a.
Next, the sprinter maintains the top speed vv for the remaining 6 seconds (since total time is 10 seconds). The distance during this phase is d2=v×6=24ad_2 = v \times 6 = 24a.
Summing these distances gives the total 100 meters: d1+d2=8a+24a=32a=100d_1 + d_2 = 8a + 24a = 32a = 100
Solving for aa gives 3.125 m/s23.125 \, m/s^2. The top speed is then 12.5 m/s12.5 \, m/s.
This approach demonstrates how constant acceleration and constant velocity phases combine to achieve a known total time and distance. It is a useful method in physics to analyze motion involving changes in velocity over time intervals, such as in sprinting or vehicle acceleration scenarios.
The sprinter covers 25 meters during acceleration and 75 meters at top speed, reflecting realistic sprint performance where acceleration is crucial in the early part of the race. Understanding these dynamics helps in training and biomechanics analysis to optimize sprinting efficiency.