Differentiation involving velocity and displacement etc..?

Question

The position of a particle moving along a straight line is given by :
s = t^3 - 6t^2 +11t -6

where s denotes the displacement in metres from a fixed point 0 at time t seconds after the start of the motion.

(i) Write down equations for velocity and acceleration at time t.

(ii) At what times is the velocity zero and what is the displacement from 0 at these times?

(iii) What is the displacement from 0 and the velocity when the acceleration is zero?

(iv) How would you plot a velocity/time graph and a displacement/time graph for t in the range 0 ≤ t ≥ 5.
i.e. how would you work out the values for the graph???

Answer 1

i) Velocity is the first derivative: 3t² - 12t + 11

Accelleration is the second derivative: 6t - 12

ii) Solve each of these for zero.

iii) Plug the answer for the accelleration at zero into the velocity equation.

iv) Plot some numbers -- i.e t = 0, t = 1, t = 2, t = 3, t = 4, and t = 5 -- on an x-y-axis, "t" being along the x-axis, and "s" being along the y-axis, and connect the dots with a smooth curve (or curves).

By the way, I think you mean "0 ≤ t ≤ 5".

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) s(t) = t^3 - 6t^2 + 11t - 6

2) By definition, the velocity, v(t), @ t is ds/dt
v(t) = 3t^2 - 12t + 11

3) By definition, the acceleration, a(t), @ t is dv/dt
a(t) = 6t - 12

4) At what times is the velocity zero?
There are two such times, T1, T2, corresponding to the zeros of v(t).

5) What is the displacement from 0 at these times? s(T1), s(T2)

6) At what time, T, is the the acceleration zero?
From 3) T = 2;

7) What is the velocity when the acceleration is zero?
v(T) = 3T^2 - 12T + 11 = -1

8) What is the displacement from 0 when the acceleration is zero?
s(T) = T^3 - 6T^2 + 11T - 6 = 0

Answer 4

(I)
the velocity is the rate of change of position in time
ds/dt = 3t^2 -12t +11
the acceleration is the rate of change of velocity in time
ds2/dt2 = 6t - 12

(ii) the velocity is zero when
3t^2 - 12t + 11 = 0
t = 12/6 +/- sqrt( 12*12 - 4*3*11)/6
= 2 +/- sqrt( 12 ) /6
= 2 +/- sqrt (1/3)
plug back into original equation for s

(iii) the acceleraion is zero when
6t-12 = 0
i.e t = 2
plug back into original equation for s, and that for ds/dt

(iv)
plug a set of values for t into original equation and that for ds/dt in the range of interest.
s should be flat at t= 2 +/- sqrt(1/3) (velocity zero)
the velocity should be flat when t=2 (accel zero)