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

1)
s = 10 + 5t +12t^2 - t^3

if you differential this displacement
it will give you velocity
that is
v= ds/dt = 0+ 5(1) + 12(2)t - 3t^2 = 5 + 24t - 3t^2
set v=0 to find the time it takes for zero velocity
0 = 5 + 24t - 3t^2
t = 8.203sec or -0.203
time is 8.203sec since t>0

differentiate v with respect to time that is
a=dv/dt=0+24(1)-3(2)t = -6(t-4)
set a = 0 therefore t=4sec
sub t=4 in
v=5 + 24t - 3t^2=5+24(4)-3(4)^2
= 53m/s


2)
T= 2pi sqrt (L/g)
sub L(new)=1.05%L=1.05
so T(new)=2pi sqrt (1.05L/g) = sqrt(1.05) x 2pi sqrt (L/g)
T(new)=sqrt(1.05) x T(old)
T(difference) = T(new) - T(old)
= sqrt(1.05) x T(old) - T(old)
= T(old) x [sqrt(1.05) -1 ]
T(percentage change) = T(diff) / T(old) *100
= 100 x [sqrt(1.05) -1 ]
= 2.47%
i help you
you help me

Answer 2

1)firstly differentiate with respect to time,i.e.ds/dt which will give the equation,speed=5+24t-3t^2.
Take speed =0 and solve it.
Again differentiate the eq^n v=5+24t-3t^2 w.r.t. time.
it will give a=24-6t and find the answers.
2)Take increase as L+(L/20)and find the answer.(Note that the time should decrease.

Answer 3

v=ds/dt =-3t^2+24t+5
3t^2-24t-5=0

t=((24+-sqrt(576+60))/6
t=8.20 s
(the - sing gives negative time)
2)dT/dL = 2pi(L/g)^-1/2*1/2g = pi/g^3/2 *1/sqrtL
dT/T= 1/(2g) dL/L and as dL/L =5%

dT/T =0.255%