The position of a particle is defined by the equation s = 0.13t3 + t2 - 3t + 2, where s is in metres and t is in seconds. This equation is valid for the time range t=0 to t=5 seconds. Determine the following:(a) the velocity of the particle at t=1.25 seconds, v = Answer m/s(b) the acceleration of the particle at t=1.25 seconds, a = Answer m/s2
Question
The position of a particle is defined by the equation s = 0.13t3 + t2 - 3t + 2, where s is in metres and t is in seconds. This equation is valid for the time range t=0 to t=5 seconds. Determine the following:(a) the velocity of the particle at t=1.25 seconds, v = Answer m/s(b) the acceleration of the particle at t=1.25 seconds, a = Answer m/s2
Solution
Sure, I can help with that.
(a) The velocity of a particle at a given time can be found by taking the derivative of the position function with respect to time.
The derivative of s = 0.13t^3 + t^2 - 3t + 2 is ds/dt = 0.39t^2 + 2t - 3.
Substituting t = 1.25 seconds into this equation gives:
v = 0.39*(1.25)^2 + 2*1.25 - 3 = -0.6125 m/s
(b) The acceleration of a particle at a given time can be found by taking the derivative of the velocity function with respect to time.
The derivative of v = 0.39t^2 + 2t - 3 is dv/dt = 0.78t + 2.
Substituting t = 1.25 seconds into this equation gives:
a = 0.78*1.25 + 2 = 2.975 m/s^2
So, the velocity of the particle at t=1.25 seconds is -0.6125 m/s and the acceleration of the particle at t=1.25 seconds is 2.975 m/s^2.
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