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) Detenine the displacement of this particle between the time interval t = 0 and
t= 1s ; t= 1s and t = 4s
c) Determine the average velocity during those two time intervals
d) Derive a general expression for the instantancous 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 = 20 - 0 + 0 = 20m
So, the initial position of the particle is 20m.
b) The displacement of the particle between t = 0 and t = 1s can be calculated by subtracting the initial position from the position at t = 1s:
Displacement = x(1) - x(0)
= (20 - 2(1) + (1)^3) - 20
= (20 - 2 + 1) - 20
= 19 - 20
= -1m
So, the displacement of the particle between t = 0 and t = 1s is -1m.
Similarly, the displacement of the particle between t = 1s and t = 4s can be calculated:
Displacement = x(4) - x(1)
= (20 - 2(4) + (4)^3) - (20 - 2(1) + (1)^3)
= (20 - 8 + 64) - (20 - 2 + 1)
= (76) - (19)
= 57m
So, the displacement of the particle between t = 1s and t = 4s is 57m.
c) The average velocity during the time interval t = 0 and t = 1s can be calculated using the formula:
Average velocity = (Displacement) / (time interval)
= (-1m) / (1s - 0s)
= -1m/s
So, the average velocity during this time interval is -1m/s.
Similarly, the average velocity during the time interval t = 1s and t = 4s can be calculated using the formula:
Average velocity = (Displacement) / (time interval)
= (57m) / (4s - 1s)
= (57m) / (3s)
= 19m/s
So, the average velocity during this time interval is 19m/s.
d) The instantaneous velocity can be found by taking the derivative of the position function with respect to time:
v(t) = d(x(t)) / dt
= d(20 - 2t + t³) / dt
= -2 + 3t²
e) To calculate the instantaneous velocity at t = 3s, we plug in t = 3 into the expression for v(t):
v(3) = -2 + 3(3)²
= -2 + 3(9)
= -2 + 27
= 25m/s
So, the instantaneous velocity at t = 3s is 25m/s.
f) The instantaneous acceleration can be found by taking the derivative of the velocity function with respect to time:
a(t) = d(v(t)) / dt
= d(-2 + 3t²) / dt
= 6t
To calculate the instantaneous acceleration at t = 2s, we plug in t = 2 into the expression for a(t):
a(2) = 6(2)
= 12m/s²
So, the instantaneous acceleration at t = 2s is 12m/s².