Example1.

A space probe is launched from a platform and moves away in straight line. An observer calls the direction of motion the x –axis and measures the distance by which the probe is displaced from the platform in term of x. Suppose then that the probe’s position at any time is given by the expression :

x(t)=A+Bt^2 wherein A=150 m and B = 30.0 m/s^2.

(a) Find the average velocity of the probe over the time interval extending from ti = 7.00 s to tf = 9.00 s
(b) What is its instantaneous velocity at time t = 8.00s


Solution



(a) We need to determine the average velocity :


vav = (xf-xi)/(tf-ti)

where xf = A+B (9.00 s)^2 and xi = A + B(7.00 s)^2
Hence
Vav = B[(9.00 s)^2 –(7.00 s)^2]/(9.00-7.00s)

and vav = (16.0 s) B = 480 m/s

Answer

(b)
From v=dx/dt = 2Bt, and so v = 2(30.0 m/s^2)(8.00 s) = 480 m/s



Answer


Next : Example 2