The position of a particle in rectilinear motion is defined by the relationship x=t3 -2t2+10t-6, where x is in metres and t is in seconds determine (i) particle’s position, velocity and acceleration at t=3secs (ii)average velocity during t=2seconds and t=3seconds(iii)average acceleration during t=2seconds and t=3seconds.?

Question

The position of a particle in rectilinear motion is defined by the relationship x=t3  -2t2+10t-6, where x is in metres and t is in seconds determine (i) particle’s position, velocity and acceleration at t=3secs (ii)average velocity during t=2seconds and t=3seconds(iii)average acceleration during t=2seconds and t=3seconds.?

Answer

Given: x = t^3 - 2t^2 + 10t - 6 (where x is in metres and t is in seconds)

(i) Particle's position, velocity and acceleration at t = 3 secs:
At t = 3 secs,
x = (3^3) - 2(3^2) + 10(3) - 6
= 27 - 18 + 30 - 6
= 33 m

To find velocity, we differentiate the given equation w.r.t time:
v = dx/dt = 3t^2 - 4t + 10
At t = 3 secs,
v = 3(3^2) - 4(3) + 10
= 23 m/s

To find acceleration, we differentiate the velocity equation w.r.t time:
a = dv/dt = 6t - 4
At t = 3 secs,
a = 6(3) - 4
= 14 m/s^2

Therefore, at t = 3 secs:
Position = 33 m
Velocity = 23 m/s
Acceleration = 14 m/s^2

(ii) Average velocity during t = 2 seconds and t = 3 seconds:
Average velocity = (Change in position) / (Change in time)
At t = 2 secs,
x = (2^3) - 2(2^2) + 10(2) - 6
= 8 m

At t = 3 secs (already calculated above),
x = 33 m

Change in position = 33 - 8 = 25 m
Change in time = 3 - 2 = 1 sec

Average velocity = 25 m/s

(iii) Average acceleration during t = 2 seconds and t = 3 seconds:
Average acceleration = (Change in velocity) / (Change in time)
At t = 2 secs,
v = 3(2^2) - 4(2) + 10
= 14 m/s

At t = 3 secs (already calculated above),
v = 23 m/s

Change in velocity = 23 - 14 = 9 m/s
Change in time = 3 - 2 = 1 sec

Average acceleration = 9 m/s^2

Therefore, at t = 2 secs and t = 3 secs:
Average velocity = 25 m/s
Average acceleration = 9 m/s^2

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