Question

A ball starts from rest at the initial position x; = 0. The
ball has a constant acceleration of 2.4 m/s2. (a) Write
the position-time equation for the ball. (b) What is the
position of the ball at t = 1.0 s? (c) What is its position
at t = 2.0 s?​

Answer 1

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

Answer 2

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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