CE AMATH --- Rate of change
2007-09-16 11:49 pm
A particle moves along the x-axis so that its displacement s at any time t(in seconds where t≧ 0) is given by s=-t^3+3t^2+15t/4+1
a) Find the rate of change of s in terms of t.]
b) Find the velocity and acceleration when the particle starts its motion.
c) Given that the particle is instantaneously at rest at point B, find the time when it happens.
回答 (1)
2007-09-16 11:55 pm
✔ 最佳答案
a) diff. the equation with respect to t i.e. -3t^2+6t+15/4b) v (at t = 0) = ds/dt (at t =0) = -3*0 + 6*0 + 15/4 = 15/4
a (at t = 0) = dv/dt (at t=0) = 6t + 6 (at t = 0) = 6
c) at rest = v = 0 = ds/dt = 0, so t = -0.5 (rejected)and 2.5, so t = 2.5
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