1) A Particle moves s metres in time t seconds where:?

Question

s = 10 + 5t +12t^2 - t^3

Find (i) the time when it's speed is zero
(showing wrkings pls)

Find (ii) the velocity when the acceleration is zero

2) The period of oscillation of a simple pendulum is given by T= 2pi sqrt (L/g) where L is the length of the pendulum and g is the acceleration due to gravity (as constant g = 9.8 m/s^2).

Find the approximate percentage change in T if L is increased by 5%.
(Pls show workings step by step)
Thanks

Answer 1

knowing s(t), v(t) is ds(t)/dt
In this case
v(t)=5+24t-3t^2
set equal to zero and solve for t
since it is quadratic there are 2 roots. The positive root is 8.2 s

a(t)=dv(t)/dt
In this case
a(t)=24-6t
set equal to zero. The time a=0 is t=4
evaluate v(4)=53

2)Isolate L
T=2pi sqrt(L)/sqrt(g)
call 2pi/sqrt(g)= n
T=n*sqrt(L)
to increase L by 5%, multiply L by 1.05
T2=n*sqrt(L*1.05)
T2/T1=sqrt(1.05)

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