[ALI ALZAHRANI]

More Examples

[1] If the position of a ball thrown vertically upward is given by the equation
y = 1.00 + 12.0t -3.00t^2 (m) .

Find the following:

(a) the average velocity in the interval [0, 2.0s],
(b) the initial velocity of the ball

Solution

(a) To find the average velocity between [0,2.0s] we have to determine the position of the ball at these times.

At t=0 we have y=1+12*0- 3*0^2=1 m

At t=2 we have y=1+12*2-3*2^2=13 m

Therefor the average velocity is

v=Dy/Dt = y2-y1/t2-t1 = 13-1/2-0 = 12/2 = 6 m/s

(b) The initial velocity means the velocity at t =0. However, to find the velocity at any time we have to find the equation of the velocity

We know that velocity is the derivative of the displacement, thus

v = dy/dt = 12 - 6t

So, the initial velocity is

v(t=0) = 12 -6*t = 12 m/s


[2] A car moves from a position of +4.0 m to a position of -1.0 m in
2.0 seconds. The initial velocity of the car is -5.0 m/s and the final
velocity is -2.0 m/s.
(a) What is the displacement of the car?
(b) What is the average velocity of the car?
(c) What is the average acceleration of the car?

Solution

(a) The displacement of the car is the total distance between the final and initial positions

Dx= x(final) - x(initial) = -1 - 4 = -5 m

(b) The average velocity of the car is obtained from

v=Dx/Dt = -5/2 = -2.5 m/s

(c) The average acceleration of the car is given by

a=Dv/Dt = v(final) - v(initial) /Dt= -2-(-5)/2 =3/2 = 1.5 m/s^2