Respuesta :
Answer:
(a) s(t) = 4.9t²
(b) s(1) = 4.9 m
(c) s(2) = 19.6 m
Explanation:
(a)
the position time equation can be formed by using 2nd equation of motion:
s = Vi t + (1/2)gt²
where,
s = position
Vi = Initial velocity = 0 m/s
t = time
g = 9.8 m/s²
Therefore,
s = (0 m/s)t + (1/2)(9.8)t²
s(t) = 4.9t²
(b)
s(1) = 4.9(1)²
s(1) = 4.9 m
(c)
s(2) = 4.9(2)²
s(2) = 19.6 m
We want to get the motion equations of a ball, and we want to use them to find the position of the ball at different times.
The solutions are:
- a) P(t) = (1.2 m/s^2)*t^2
- b) x = 1.2m
- c) x = 4.8m
We know that the ball starts from rest (so the initial velocity is zero) from the position x = 0.
The ball has a constant acceleration of 2.4 m/s^2
Then the acceleration equation is:
A(t) = 2.4 m/s^2
To get the velocity equation we integrate, remember, the initial velocity is zero, so we don't have a constant of integration:
V(t) = (2.4 m/s^2)*t
To get the position equation we integrate again, and because the initial position is x = 0, we don't have a constant of integration:
P(t) = 0.5*(2.4 m/s^2)*t^2
a) The position equation is:
P(t) = (1.2 m/s^2)*t^2
b) The position of the ball at t = 1s is given by:
P(1s) = (1.2 m/s^2)*(1s)^2 = 1.2m
The ball position at t = 1s is x = 1.2m
c) Similar to the above case:
P(2s) = (1.2 m/s^2)*(2s)^2 = 4.8 m
x = 4.8m
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