How far does a cyclist travel when she accelerates from rest to 6.5m/s in 22s?
V = Vo + a*t
V = 6.5 m/s
Vo = 0
t = 22 s.
Solve for a.
d = 0.5a*t^2
t = 22 s.
Solve for d
143
To find how far the cyclist travels, we can use the equations of motion. The equation that relates displacement (distance), initial velocity, acceleration, and time is:
d = vt + (1/2)at^2
Where:
d = displacement (distance)
v = initial velocity
a = acceleration
t = time
In this case, the cyclist starts from rest (initial velocity, v = 0 m/s) and accelerates to 6.5 m/s in 22 seconds. Therefore, the initial velocity is 0 m/s, the final velocity (6.5 m/s) is the velocity after accelerating for 22 seconds, and the acceleration is the rate at which the velocity changes over time.
Plugging in the values in the equation:
d = (0 m/s)(22 s) + (1/2)(a)(22 s)^2
Since we are trying to find the distance, let's solve for d by substituting the given information:
d = (1/2)(a)(22 s)^2
Now, we need to find the acceleration (a) of the cyclist. To do that, we can use the equation:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Since the cyclist starts from rest (u = 0 m/s) and accelerates to 6.5 m/s in 22 seconds, we can solve for a:
6.5 m/s = 0 m/s + a(22 s)
6.5 m/s = 22 a s
a = 6.5 m/s / 22 s
a ≈ 0.30 m/s^2
Now, we can substitute the value of a back into the equation for distance:
d = (1/2)(0.30 m/s^2)(22 s)^2
d = (0.15)(22^2) m
d ≈ 72.6 m
Therefore, the cyclist travels approximately 72.6 meters when she accelerates from rest to 6.5 m/s in 22 seconds.