(i) Find an expression for the velocity v(t) and acceleration a(t) of the particle in terms of t.

v(t) = 3t² - 24t +36

a(t) = 6t - 24

(ii) When is the particle at rest?

At rest when v = 0.

∴ 3(t² - 8t + 12) = 0

(t - 2)(t - 6) = 0

At rest when t = 2 mins or t = 6 mins. 

(iii) When does the particle reach its maximum velocity in the negative direction?

Maximum velocity when acceleration = 0.
So at t = 4 mins.

(iv) Complete the table:.

t	s(t)	v(t)	a(t)
0	0	36	-24
2	32	0	-12
4	16	-12	0
6	0	0	12
8	32	36	24

 

(v) Sketch all 3 graphs and describe the motion of the particle in the first 8 seconds.

• The particle starts at the origin with a velocity of 36 cm/sec and an acceleration of -14 cm/min².
• during the first two minutes, the particle moves with a decreasing but positive velocity and a negative acceleration. At t = 2, the velocity is zero and so the particle is at the max distance away (at 32 cm).
• at t = 4, the acceleration is zero and so the particle is at minimum velocity of -12 cm/min.
• from t = 4 to t = 6, the particle has positive acceleration so the particle slows down in the negative direction until it is at rest at t = 6 at the origin.
• from t = 6 to t = 8, the particle still has positive acceleration so its speed increases from rest in the positive direction and it moves away in the positive direction to 32 cm.

(i) Find the initial velocity and acceleration particle.

Vel = 6t - 3t²

Accel = 6 - 6t

At t = 0, vel = 0 and accel = 6 m/sec².

(ii) Find the stationary points on the displacement-time graph and state their nature.

SP points when vel = 0.

3t(2- t) = 0 so t = 0 secs or t = 2 secs.

At t = 0, a = 6 > 0 so a (local) min displacement of 0.

At t = 2, a = -6 < 0 so a max displacement of 4.

(iii) Determine the time at which the acceleration of the particle is zero.

Acceleration = 6 - 6t = 0 - so at t = 1.

(iv) Graph the function x(t), t ≥ 0.

(v) Determine the total distance travelled by the particle in the first 4 seconds.

The particle moves from the origin (t = 0) to 4 (t = 2) and then back to the origin (t = 3).

At t = 4, the particle is at 3(4)² - 4³ = -16 m.

So total distance travelled is 4 + 4 + 16 = 24 m.

(vi) Describe the motion of the particle.

• the particle starts from the origin at t = 0 from rest with a positive acceleration.
• at t = 2, the particle is at its maximum positive displacement of 4, at rest and with a negative acceleration.
• at t = 3, the particle is back at the origin and travelling with a velocity of -9 m/sec and still with a negative acceleration.
• at t = 4, th particle is at 24 m to the left of the origin and travelling with increased negative velocity and acceleration (forever after!!).

(ii) When is the particle at rest?

(iii) What is the acceleration of the particle when it is first at rest?

(iv) Find the distance travelled by the particle in the first three seconds.

As the particle was at rest at t = 2, it changed direction and started coming back (velocity at 2.5 secs = -0.75 m/sec). Hence the calculation of distance must be done in 2 parts:

T = 0, x = 2 m (starting point).

T = 2, x = 16 m - so travelled a distance of 14 m.

T = 3, x = 15.5 m - so travelled backwards 0.5 m.

Total distance = 14.5 m in the first 3 seconds.

(easier to expand now).

(i) Find the initial position of the body.

t = 0, x = 0 so at the origin.

(ii) When is the body again at this position?

When x = 0, t = 0 or 6. So at the origin again at x = 6 seconds.

(iii) Obtain the velocity and accelerations functions for the motion.

Velocity = 3t² - 24t + 36

Accel = 6t - 24

(iv) At what times is the body momentarily at rest?

When velocity = 0

Velocity = 3(t² - 8t+12) = 3(t - 2)(t - 6)

∴ at rest at t = 2 and t = 6 seconds.

(v) In which direction does the body move between these times?

Try t = 3, v = -9 so body moves in a negative direction (i.e. towards the left).

Rather than try t = 3, we could have said for values between 2 and 6, velocity is negative by looking at the factored equation.

(vi) What is the initial acceleration for the motion? Is it tending to slow down or speed up the initial movement?

t = 0, acceleration = -24 m/sec².

The negative acceleration is slowing down the motion which started at 36 m/sec. Hence it stops at 2 secs.

(vii) At what instant is there no acceleration acting on the moving body? What is the velocity at that instant? Show that this is the greatest speed in a negative direction attained by the moving body.

When accel = 0, 6t - 24 = 0 so t = 4 secs.

At t = 4, velocity = -12 m/sec.

The form of the velocity equation is a concave up parabola with zeros at t = 2 and t = 6. Hence the maximum speed in the negative direction is reached half way between those times.

(viii) Draw graphs to show each of the displacement, velocity and acceleration functions.

(i)

(ii) When at rest, v = 0

∴ 3t(60 - t) = 0

t = 60 secs

(iii) When t = 60, h = 276 m.

(iv) When the lift is travelling fastest, acceleration = 0.

(v)

When t = 0, v = 2(2 - 0)e⁰ = 4 m/sec.

(ii) When and where will the particle be at rest?

When v = 0, t = 2 secs (as the exponential component can't = 0).

(iii) Find the time when the particle has zero acceleration.

(iv) What is the time interval when the particle is increasing its speed?

Speed increases either when vel > 0 and accel > 0

OR when vel < 0 and accel < 0.

Here speed > 0 up to t = 2 and then it becomes -ve.

Accel is -ve up to t = 4.

Up to 2 seconds, speed > 0 and accel < 0 - so no increase.

The only interval is therefore 2 < t < 4 because here both speed < 0 and a < 0.

Difficult.

(i) At what times during the period 0 < t < 3π is the particle stationary?

vel = 1 + cos t = 0

cos t = -1

t = π (as 3π is excluded from the domain).

(ii) At what times during the period 0 < t < 3π is the acceleration equal to zero?

accel = - sin t = 0

t = π and 2π (0 and 3π and excluded from the domain).

(iii) Sketch the graph x = t + sin t for 0 < t < 3π. Clearly label any stationary points and any points of inflexion.

(ii) & (iii)

(iv) Displacement = 18 m to the left.

(i) Find the initial acceleration of the object.

(ii) Find when the object is at rest.

(iii) Find the minimum distance between the origin and the object during its motion.

Answer.(i) Init accel - 14 m/sec².
(ii) At rest after 2/3 or 4 seconds.
(iii) Min distance = 14 m.

The equation for velocity is v = (t - 2)² + 5. The first term must be positive (it is a square) then 5 is added. When the first term is 0 at t = 2, minimum velocity must be 5 m/sec.

(ii) Find the initial acceleration of the particle.

a = 2(t - 2) so when t = 0, acceleration is -4 m/sec².

(iii) Find the distance travelled by the particle in the first 3 seconds.

The velocity is always positive, so there is no problem about calculating distance in two parts. Just integrate because the area under the velocity curve is the distance travelled.

(iv) Initially the particle is at 9 m to the left of the origin. Find the position of the particle at t = 5 secs.

20. (i)

(ii)

(iii)

21. (i)

(ii)

Distance travelled must be split into two parts:

Distance from t = 0 to t = π/12 and then to t = π/3:

(iii)