The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29

The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29: Which of the following formulae are correct for describing the motion of the particle over the time-interval t₂ to t₁?

 

(a) x(t₂) = x(t₁) + v(t₁)(t₂-t₁) + (1/2)a(t₂-t₁)²

(b) v(t₂)= v(t₁) + a(t₂-t₁)

(c) v_Average = (x(t₂) - x(t₁)) / (t₂ - t₁)

(d) a_Average = (v(t₂) - v(t₁)) / (t₂ - t₁)

(e) x(t₂) = x(t₁) + v_Average(t₂ - t₁) + (1/2       )a_Average(t₂ - t₁)²

(f) x(t₂) - x(t₁) = area under the v-t curve bounded by the t-axis and the dotted line shown.

Answer:

The correct formulae describing the motion of the particle are (c), (d) and, (f)

The given graph has a non-uniform slope. Hence, the formulae given in (a), (b), and (e) cannot describe the motion of the particle. Only relations given in (c), (d), and (f) are correct equations of motion.

Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m/s and 30 m/s

Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m/s and 30 m/s. Verify that the graph shown in Fig. 3.27 correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not  rebound after hitting the ground. Take g = 10 m/s2. Give the equations for the linear and curved parts of the plot.

Answer:

For first stone:

Initial velocity, uI = 15 m/s

Acceleration, a = -g = - 10 m/s²
Using the relation,

When this stone hits the ground, x₁ = 0
→- 5t² + 15t + 200 = 0

t² - 3t - 40 = 0

t² - 8t + 5t - 40 = 0

t (t - 8) + 5 (t - 8) = 0
t = 8 s or t = - 5 s

Since the stone was projected at time t = 0, the negative sign before time is meaningless.

→t = 8 s

For second stone:

Initial velocity, uII = 30 m/s

Acceleration, a = -g = - 10 m/s²
Using the relation,

At the moment when this stone hits the ground; x₂ = 0 - 5t2 + 30 t + 200 = 0

t2 - 6t - 40 = 0

t2 - 10t + 4t + 40 = 0

t (t - 10) + 4 (t - 10) = 0
t (t - 10) (t + 4) = 0
t = 10 s or t = - 4 s

Here again, the negative sign is meaningless. →t = 10 s

Subtracting equations (i) and (ii), we get

Equation (iii) represents the linear path of both stones. Due to this linear relation between (x₂ - x₁) and t, the path remains a straight line till 8 s.
Maximum separation between the two stones is at t = 8 s.
(x₂ - x₁)max = 15× 8 = 120 m

This is in accordance with the given graph.

After 8 s, only second stone is in motion whose variation with time is given by the quadratic equation:

x₂ - x₁ = 200 + 30t - 5t2

Hence, the equation of linear and curved path is given by x₂ - x₁ = 15t (Linear path)

x₂ - x₁ = 200 + 30t - 5t2 (Curved path)

On a long horizontally moving belt (Fig. 3.26)

On a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed 9 km h^-1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h^-1. For an observer on a stationary platform outside, what is the

(a) speed of the child running in the direction of motion of the belt ?.

(b) speed of the child running opposite to the direction of motion of the belt ?

(c) time taken by the child in (a) and (b) ?

Which of the answers alter if motion is viewed by one of the parents?

 

Answer:

(a) Speed of the belt, vB = 4 km/h
Speed of the boy, vb = 9 km/h

Since the boy is running in the same direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as:
vbB = vb + vB = 9 + 4 = 13 km/h

(b) Since the boy is running in the direction opposite to the direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as:
vbB = vb + (- vB) = 9 - 4 = 5 km/h

(c) Distance between the child’s parents = 50 m

As both parents are standing on the moving belt, the speed of the child in either direction as observed by the parents will remain the same i.e., 9 km/h = 2.5 m/s.

Hence, the time taken by the child to move towards one of his parents is

(d) If the motion is viewed by any one of the parents, answers obtained in (a) and

(b) get altered. This is because the child and his parents are standing on the same belt and hence, are equally affected by the motion of the belt. Therefore, for both parents (irrespective of the direction of motion) the speed of the child remains the same i.e., 9 km/h.

For this reason, it can be concluded that the time taken by the child to reach any one of his parents remains unaltered.