10. A particle moves along the x-axis component varies with time according to equation
given below.
where t is in Second.
x(t) = (20- 2t + t³)m
a) Determine initial position of the particle.
b) Determine the displacement of this particle between the time interval t =0 and
t= 1s; t= 1 and t = 4s
c) Determine the average velocity during those two time intervals
d) Derive a general expression for the instantaneous velocity as a function of time.
e) calculate instantaneous velocity at t = 3s.
f) compute the instantaneous acceleration at t = 2s
a) The initial position of the particle can be determined by plugging in t = 0 into the equation for x(t).
x(0) = (20 - 2(0) + (0)³) = 20m
b) The displacement of the particle between t = 0 and t = 1s can be found by calculating the difference in x values at these two times.
Displacement = x(1) - x(0) = (20 - 2(1) + (1)³) - 20 = 19m - 20m = -1m
The displacement of the particle between t = 1s and t = 4s can be found in the same way.
Displacement = x(4) - x(1) = (20 - 2(4) + (4)³) - (20 - 2(1) + (1)³) = 68m - 19m = 49m
c) The average velocity during the first time interval can be found by dividing the displacement by the time interval.
Average velocity = Displacement / (1 - 0) = -1m/1s = -1m/s
The average velocity during the second time interval can be found in the same way.
Average velocity = Displacement / (4 - 1) = 49m/3s = 16.33m/s
d) The instantaneous velocity can be found by taking the derivative of the equation for x(t) with respect to time.
v(t) = d/dt (20 - 2t + t³)
v(t) = -2 + 3t²
e) To find the instantaneous velocity at t = 3s, plug t = 3 into the expression for v(t).
v(3) = -2 + 3(3)² = -2 + 27 = 25m/s
f) The instantaneous acceleration can be found by taking the derivative of the equation for v(t) with respect to time.
a(t) = d/dt (-2 + 3t²)
a(t) = 6t
To find the instantaneous acceleration at t = 2s, plug t = 2 into the expression for a(t).
a(2) = 6(2) = 12m/s²