The motion of a particle is defined by the relation x = t3 − 9t2 + 24t − 6, where x is expressed in metres and t in second. Determine the position, velocity, and acceleration when t = 5s.
Answers
=18−6t
The time will be
0=18−6t
t=3 sec
The position of the particle at 3 sec
Put the value of t in equation (I)
x=9×9−27
x=81−27
x=54 m
The position of the particle will be 54 m.
Hence, Option A is correct
Answer:
The acceleration of the particle when t = 5s is 12 m/s^2.
Explanation:
Given, x = t^3 - 9t^2 + 24t - 6
To find the position when t = 5s, substitute t = 5 in the given equation:
x = (5)^3 - 9(5)^2 + 24(5) - 6
x = 125 - 225 + 120 - 6
x = - (225 - 120 - 6)
x = -99 meters
Therefore, the position of the particle when t = 5s is -99 meters.
To find the velocity, differentiate x with respect to t:
v = dx/dt = 3t^2 - 18t + 24
Substitute t = 5s to get the velocity at t = 5s:
v = 3(5)^2 - 18(5) + 24
v = 75 - 90 + 24
v = 9 m/s (approx.)
Therefore, the velocity of the particle when t = 5s is 9 m/s (approx.).
To find the acceleration, differentiate v with respect to t:
a = dv/dt = 6t - 18
Substitute t = 5s to get the acceleration at t = 5s:
a = 6(5) - 18
a = 12 m/s^2
Therefore, the acceleration of the particle when t = 5s is 12 m/s^2.
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