Starting from rest, a particle is moving in a straight line and has an accelerat

Question

Starting from rest, a particle is moving in a straight line and has an acceleration ot a = (2t - 6) m/s^2, where t is in seconds. Now answer the following questions: a. At t = 3 s, the velocity of the particle can be said constant, true or false? b. Write down the range of time when the particle will have deceleration. c. What is the particle's velocity (m/s) when t = 6 s? Ans: (i) 0 (ii) 5 (iii) 10 (iv) 15 (v) 20 d. What is its position (m Approx.) when t = 11 s? Ans: (i) 71 (ii) 81 (iii) 91 (iv) 101 (v) 121

Explanation / Answer

a = dv/dt
dv = a dt
dv = (2t - 6) dt
Integratrating both sides with respect to t
v(t) = t^2 - 6t + C

Initial condition, starts from rest, v(0) = 0
0 = 0^2 - 0 + C
C = 0

v(t) = t^2 - 6t

at t = 3 sec

v(t) = 9 - 18 = -9 m/s particle is deaccelerate.

range of particle is 0 to -9

v(t) = 6^2 - 6x6 = 0 ------at t = 6 sec

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