Particles moves in straight line so that displacement, s metres at time, t seconds from starting point at t=0?

Question

is given by s=2t+3t^2-t^3
a)differentiate so to find velocity and aceleration
b)what is speed at t=0.5 seconds
c)find all stationary points in displacement function
d)when does velocity reach maximum and what is maximum

Please explain all working if poss - not just answers

Answer 1

a) First derivative gives velocity:
v = 2 + 6t - 3t^2
Second derivative gives accaeleration:
a = 6 - 6t

b) v(0.5) = 2 + 6*0.5 -3*0.5^2 = 4.25 m/s

c) s = t(2 +3t -t^2)
Thus at s = 0, either t = 0 or 2 +3t -t^2 = 0.
This latter has solutions t = (3+/-sqrt(17))/2

d) The velocity reaches its maximum when a = 0, that is, at t = 1.
This maximum is v = 2 + 6 - 3 = 5 m/s

Answer 2

Position of particle along straight line (s) s = t^3 - 6t^2 +11t -6 .........(1) (a) for velocity, we differentiate the displacement S, as velocity at any instant is rate of change of displacement. so V (t)= ds/dt = 3 t^2 - 12 t + 11 ----(2) for knowing the acceleration at time t, we differentiate the velocity, as acceleration at any instant is rate of change of velocity, so f(t) = dV(t) /dt = 6 t - 12 ----(3) (b) velocity is zero at two times 3 t^2 - 12 t + 11 =0 >> t1 and t2 sqtr - values ************ I missed velovity = 0 part******* calculate as did in f(t) = 0 part now the time at which acceleration is zero is when f(t) = 0 = 6 t - 12 or t = 2 sec after 2 sec from start its acceleration becomes zero. (c) when acceleration becomes zero , i.e t=2, its displacement from point o is obtained by putting t=2 in equ (1) is at point 0 S(0) = -6 s(2) = (2)^3 - 6(2)^2 +11(2) -6 = 8 - 24 + 22 - 6 = 0 displacement from t=2 to t=o = S(2) - S(0) = +6 meter Velocity: velocity at t=0 V (0)= 11 (point o) clock started V(2) = 3*4 - 12*2 + 11 = - 1 m/s so its relative velocity V (f=0 and point 0) = - 1 - 11 = -12 (d) you take equ (1) and (2) S-T graph: mark t=o value S(o) - this point will be on - S axis/ now put the values T = 1 2 3 4 5 in equ (1) and find out S(1 to 5) separately and mark on the graph. Now join them in sequence of T. V-T graph: mark t=o value V(o) - this point will be on + V axis/ now put the values T = 1 2 3 4 5 in equ (2) and find out V(1 to 5) separately and mark on the graph. Now join them in sequence of T. from V_T if you calculate the area of shape formed it will give give S. as dS =V dT

Answer 3

1)

v(t) = 2 +6t -3t^2
a(t) = 6 - 6t

2) v(.5) = 2 + 6(.5) -3(.5)^2 = 4.25 m/s

3) stationary points of the displacement will be when the derivative of the displacement is zero. that is solve v(t) =0 and plug t's into the displacement to find the y's.

i.e. 2 +6t -3t^2 = 0

4) Vmax @ t =1 --> v(1) = 5 m/s

Answer 4

differentiate first for velocity (1st order differential)

a) ds / dt = 2 + 6t - 3t^2 (this is velocity)

d^2S / dt^2 = 6 - 6t (this is acceleration and is 2nd order differential)

Speed uses top equation (velocity)

Therefore-: b) 2 + (6 x 0.5) - 3 x (0.5^2) = 4.25 ms^-1

c) Set ds / dt = 0 and then solve for t

d) Velocity is max after 1 second and is 5 ms^-1

Answer 5

a)
velocity v=ds/dt=2+6t-3t^2
acceleration a=dv/dt=d^2s/dt^2
=(6-6t)msec^(-2)

b)plug t=0.5 into v=2+6t-3t^2
v=2+6*0.5-3*(0.5)^2
=2+3-0.75=4.25msec^(-1)

c)the stationary points are at
v=0
that is,2+6t-3t^2=0, (multiply by -1)
3t^2-6t-2=0
use the quadratic formula;
t=(6+/-sqrt(36+24))/6
=1+/-sqrt(36[5/3])/6
={1+/-sqrt(5/3)}sec

at t=1+sqrt(5/3),
s=t(2+3t-t^2)
=(1+sqrt(5/3))[2+3(1+sqrt(5/3)
-(1+sqrt(5/3))^2]
do the working
=2[2+(5/3)sqrt(5/3)]m

at t=(1-sqrt(5/3)
s=(1-sqrt(5/3))
[2+3(1-sqrt(5/3)
-(1-sqrt(5/3))^2]
do the working
=2[2-(5/3)sqrt(5/3)]m

therefore,stationary points are;

1+sqrt(5/3),
2[2+(5/3)sqrt(5/3)]
{max} and,
(1-sqrt(5/3),
2[2-(5/3)sqrt(5/3)]
{min}

[at t=1+sqrt(5/3),
a=6-6t is -ve,making s a max
at t=1-sqrt(5/3),
a=6-6t is +ve, making s a min]


d)the max velocity V is at
a=0,that is,6-6t=0,>>t=1
therefore,
V=2+6t-3t^2 {t=1}
=2+6-3=5msec^(-1)

i hope that this helps