At the end of a race a runner decelerates from a velocity of 7.00 m/s at a rate of 0.400 m/s^2.
(a) How far does she travel in the next 15.0 s
At the end of a race a runner decelerates from a velocity of 7.00 m/s at a rate of 0.300 m/s2.
b) What is her final velocity?
x=vt+(1/2)*a*(t)^2
x=7*15-(1/2)*4*(15)^2
A helicopter blade spins at 110 revolutions per minute. Its tip is 5.50 m from the center of rotation.
To find how far the runner travels in the next 15.0 s, we need to use the equations of motion. In this case, we have the initial velocity (7.00 m/s), the acceleration (deceleration) (-0.400 m/s^2), and the time (15.0 s).
The equation we can use here is:
d = v₀t + (1/2)at²
Where:
- d is the distance traveled
- v₀ is the initial velocity
- t is the time
- a is the acceleration (deceleration in this case)
Now let's plug in the given values and calculate the distance:
d = (7.00 m/s)(15.0 s) + (1/2)(-0.400 m/s^2)(15.0 s)²
First, we calculate (1/2)(-0.400 m/s^2)(15.0 s)²:
(1/2)(-0.400 m/s^2)(15.0 s)² = (1/2)(-0.400 m/s^2)(225.0 s²) = -45.0 m
Now, we calculate (7.00 m/s)(15.0 s):
(7.00 m/s)(15.0 s) = 105.0 m
Finally, we add the two results together:
d = 105.0 m + (-45.0 m) = 60.0 m
Therefore, the runner travels a distance of 60.0 meters in the next 15.0 seconds.