A cheetah starts from rest and accelerates at 3.50m/s^2
a) Calculate the velocity of the cheetah ad t = 4.00s
b) Calculate the position of the cheetah at t = 4.00s when initial position is 0
To answer these questions, we can utilize the equations of motion under constant acceleration.
a) To calculate the velocity of the cheetah at t = 4.00s, we can use the equation:
v = u + at
where:
v = final velocity of the cheetah
u = initial velocity of the cheetah (which is 0, as it starts from rest)
a = acceleration of the cheetah (given as 3.50 m/s^2)
t = time (which is 4.00s)
Plugging in the values, we get:
v = 0 + (3.50 m/s^2) * (4.00s)
v = 14.00 m/s
Therefore, the velocity of the cheetah at t = 4.00s is 14.00 m/s.
b) To calculate the position of the cheetah at t = 4.00s when the initial position is 0, we use the equation:
s = ut + (1/2)at^2
where:
s = displacement or position of the cheetah
u = initial velocity of the cheetah (0 m/s)
a = acceleration of the cheetah (3.50 m/s^2)
t = time (4.00s)
Plugging in the values, we get:
s = (0 m/s) * (4.00s) + (1/2) * (3.50 m/s^2) * (4.00s)^2
s = (0 m) + (1/2) * (3.50 m/s^2) * (16.00s^2)
s = 28.00 m
Therefore, the position of the cheetah at t = 4.00s, when the initial position is 0, is 28.00 m.
To calculate the velocity of the cheetah at time t = 4.00s, we can use the formula for constant acceleration:
v = u + at
where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time
In this case, the cheetah starts from rest, so its initial velocity is 0 m/s. The acceleration is 3.50 m/s^2, and the time is 4.00s.
Plugging these values into the formula, we have:
v = 0 + (3.50)(4.00)
v = 14.00 m/s
Therefore, the velocity of the cheetah at t = 4.00s is 14.00 m/s.
To calculate the position of the cheetah at t = 4.00s when its initial position is 0, we can use another formula for constant acceleration:
s = ut + 0.5at^2
where:
- s is the position
- u is the initial velocity
- a is the acceleration
- t is the time
In this case, the cheetah starts from rest, so its initial velocity is 0 m/s. The acceleration is still 3.50 m/s^2, and the time is 4.00s.
Plugging these values into the formula, we have:
s = 0(4.00) + 0.5(3.50)(4.00)^2
s = 0 + 0.5(3.50)(16.00)
s = 0 + 0.5(56.00)
s = 0 + 28.00
s = 28.00 m
Therefore, the position of the cheetah at t = 4.00s, when its initial position is 0, is 28.00 m.