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.
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"
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