Kinematics

Nathpoop SG

A particle moves in a straight line so that its displacement, s cm, from a fixed point O, t seconds after passing A is given by s = t^3 + 12t^2 - 540t - 5. Calculate,

(a) the value of t when the particle is instantaneously at rest,

*I solved the value of t which is 10s.

(b) the velocity of the particle when it returns to the starting point.

*I can't seem to find the answer or how to do.

*Don't know how to do.

Pls help me in part (b) and (c).

THANKS!

kench-an SG

a.) correct, t = 10s

b.) look at the eqn again. if you let t = 0 (when object just passes A, NOT when object leaves O as stated in the qn), s = -5. That only means that A is 5 m away from O. So, the real displacement equation FROM START POINT O is just s = t^3+12t^2-540t. Letting s = 0 and kicking out the trivial sol of t=0, just solve the quadratic equation of t^2+12t-540 = 0 ==> t = 18 or -30

Hence, the object returns to O at t=18s

Then, find v = ds/dt and sub in t = 18 to find v when object is back at O

c.) Total dist travelled = distance from start to instantaneous rest (=dist A) + dist from instantaneous rest back to O again (=dist B)

dist A = s at t=10 minus s at t=0

dist B = s at t=18 minus s at t=10 (find absolute value, otherwise u get negative answer)

dist A + (absolute value of) dist B = game over

Nathpoop SG

Thanks so much, but what do you mean by "kicking out the trivial sol of t=0, just solve the quadratic equation of t^2+12t-540 = 0" ?

Thanks!

kench-an SG

if you try to solve t^3+12t^2-540^t = 0, can divide both side by t or rather, cancel t from both side. or rather, if you factorize, one solution is t=0 (called trivial solution, bcos ans is 0)