A biker starts moving from rest with a constant acceleration of 2.0 m/s^2. What is his velocity after 5.0 s?
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Angular velocity of the rotating tires can be calculated using the formula,
v=ω×r
Here, v is the velocity of the tires = 32 m/s
r is the radius of the tires= 0.42 m
ω is the angular velocity
Substituting the values we get,
32=ω×0.42
ω= 32/0.42 = 76.19 rad/s
= 76.19× revolution per min
=728 rpm
Angular velocity of the rotating tires is 76.19 rad/s or 728 rpm.
a) 2.64 s
We can solve this part of the problem by using the following SUVAT equation:
where
s is the displacement of the stone
u is the initial velocity
t is the time
a is the acceleration
We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:
- s (displacement) negative, since it is downward: so s = -75.0 m
- u (initial velocity) positive, since it is upward: +15.5 m/s
- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)
Substituting into the equation,
Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.
b) 10.4 m/s
The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:
where again, we must be careful to the signs of the various quantities:
- u (initial velocity) positive, since it is upward: +15.5 m/s
- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2
Substituting t = 2.64 s, we find the final velocity of the stone:
where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.
c) 4.11 s
In this case, we can use again the equation:
where
s is the displacement of the package
u is the initial velocity
t is the time
a is the acceleration
We have:
s = -105 m (vertical displacement of the package, downward so negative)
u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)
a = g = -9.8 m/s^2
Substituting into the equation,
Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after
t = 4.11 seconds.