Question

The position of an object moving along a straight line is given by x = 3 − 2t

2 + 4t

3 where x is

in meters and t in seconds.

a) Derive the expressions for the velocity and acceleration of the object as a function of time.

b) Find the position of the object at t = 0, t = 2s, t = 4s.

c) Find the displacement or the object between t = 2s and t = 4s; between t = 0s and t = 4s.

d) Find the average velocity between t = 2s and t = 4s; between t = 0s and t = 4s;

between t = 1s and t = 3s.

e) What is the instantaneous velocity at t = 2s? at t = 5s?

f) At what time(s) is/are the instantaneous velocities zero?

g) When does the instantaneous velocity have a maximum or a minimum value?

h) Find the change in velocity between t = 2s and t = 5s.

i) Find the average acceleration between t = 2s and t = 5s; between t = 1s and t = 3s.

j) Find the instantaneous acceleration of the object at t = 2s; t = 5s.

Answer 1

Position is given by, $x=3-2t^2+4t^3$

Velocity is given by, $v=\frac{dx}{dt}=-4t+12t^2$

Acceleration is given by, $a=\frac{dv}{dt}=-4+24t$

Put value of t to find the position at any time,

Position at t=0, $x=3m$

Position at t=2, x=27m

Position at t=4, x=227m

Displacement is difference of final position and initial position.

$x(t=2-t=4)=227-27=200m$

$x(t=0-t=4)=227-3=224m$

Average velocity is total displacement per unit total time.

$v(t=4-t=2)=(200)/2=100m/s$

$v(t=0-t=4)=(224)/4=55m/s$

$v(t=3-t=1)=(93-5)/2=44m/s$

Velocity at any time is given by simply putting value of time.

$v(t=2)=40m/s$

$v(t=5)=280m/s$

Time when velocity is zero, $-4t+12t^2=0 \implies t=0, 1/3s$

For velocity to be maximum or minimum, first derivative of the velocity must be zero.

$-4+24t=0 \implies t=6s$

Change in velocity in t=2 - t=5 s, $v=280-40=240m/s$

Average acceleration is total change in velocity per unit total time.

$a(t=2-t=5)=(240)/3=80m/s^2$

$a(t=1-t=3)=(96-8)/(3-1)=44m/s^2$

Instantaneous acceleration is given by simply putting value of time.

$a(t=2)=48-4=44m/s^2$

$a(t=5)=120-4=116m/s^2$