A sprinter explodes out of the starting block with an acceleration of +2.79 m/s2 , which she sustains for 1.65 s. Then her acceleration drops to zero for the rest of the race. What is her velocity at t = 1.65 s and at the end of the race?
For the velocity at t = 1.65 s, use
v = a t = 4.60 m/s.
Since there is no acceleration or deceleration after that, 4.6 m/s remains the velocity.
To find the velocity of the sprinter at different times during the race, we need to use the equations of motion. We'll break down the problem into two parts: the initial acceleration phase and the remaining time when acceleration is zero.
1. Initial acceleration phase:
During this phase, the sprinter has an acceleration of +2.79 m/s^2 for a time of 1.65 s. We can use the equation:
v = u + at
where:
v = final velocity
u = initial velocity (which is usually taken as 0 for simplicity)
a = acceleration
t = time
Plugging in the values:
u = 0 m/s (since the sprinter starts from rest)
a = +2.79 m/s^2 (positive because the acceleration is in the same direction as velocity)
t = 1.65 s
v = 0 + (2.79 m/s^2) * (1.65 s)
v = 4.5985 m/s
So, at t = 1.65 s, the sprinter has a velocity of +4.5985 m/s.
2. Remaining time with zero acceleration:
After the initial acceleration phase, the sprinter continues to run with a constant velocity (no acceleration). Since there is no change in velocity, the final velocity during this period will be the same as the velocity at t = 1.65 s.
Therefore, at the end of the race, the sprinter's velocity remains constant at +4.5985 m/s.