DC Pandey Solutions: Motion in One Dimension - 2

Introductory Exercise 3.2

Ques 1: A ball is thrown vertically upwards. Which quantity remains constant among, speed, kinetic energy, velocity and acceleration?
Ans: acceleration
Sol: Acceleration remains constant (ignoring air resistance) and is equal to g, the acceleration due to gravity, directed downwards. Speed and kinetic energy change as the ball rises and falls; velocity changes sign at the highest point while acceleration remains the same throughout the motion.
Ques 2: Equation

 does not seem dimensionally correct, why?

Ans: 

Sol: 

 The expression s_t denotes the displacement in the t^th second and is defined as the difference between displacements at times t and (t - 1):

s_t = s(t) - s(t - 1).

If the right-hand side of the given equation mixes quantities having different dimensions (for example, a plain length with a quantity that has dimensions of length × time), the equality is dimensionally inconsistent. Using the definition above shows that the intended formula must refer to a difference of two displacements; any form that does not respect this difference is dimensionally incorrect.

Ques 3: Can the speed of a particle increase as its acceleration decreases? If yes give an example.

Ans: Yes, in simple harmonic motion

Sol: Yes. Consider simple harmonic motion (SHM). As the particle moves from the extreme position towards the mean position, its speed increases while the magnitude of acceleration (which is proportional to displacement from the mean) decreases. Thus speed can increase even though acceleration decreases in magnitude. This happens because acceleration in SHM depends on displacement, not directly on speed.

Ques 4: The velocity of a particle moving in a straight line is directly proportional to 3/4th power of time elapsed. How does its displacement and acceleration depend on time?

Ans:  t^7/4, t^-1/4

Sol: v ∝ t^3/4 (given).

  ...(i)

Integrating velocity to get displacement:

or

Therefore s ∝ t^7/4 (displacement grows as t^7/4).

Differentiating v with respect to time to get acceleration:

Hence a ∝ t^-1/4 (acceleration decreases with time as t^-1/4).

Ques 5: A particle is projected vertically upwards with an initial velocity of 40 m/s. Find the displacement and distance covered by the particle in 6 seconds. Take g = 10 m/s².

Ans: 60 m, 100 m

Sol: Displacement after t = 6 s is given by s = v₀t - ½gt².

= 40 × 6 - ½ × 10 × 6² = 240 - 180 = 60 m upward.

To find the distance travelled in 6 s note that time to reach maximum height is t_max = v₀/g = 40/10 = 4 s.

The maximum height reached is h = v₀²/(2g) = 40²/(2×10) = 80 m.

In the first 4 s the particle travels 80 m upward. In the next 2 s it descends from the top; the distance descended in 2 s from rest at the top is ½ g (2)² = 20 m. Thus total distance covered in 6 s = 80 + 20 = 100 m.

Ques 6: Velocity of a particle moving along positive x-direction is v = (40 - 10t) m/s. Here, t is in seconds. At time t = 0, the x coordinate of particle is zero. Find the time when the particle is at a distance of 60 m from origin.

Ans: 2 s, 6 s, 2 (2 + √7) s

Sol: v = 40 - 10t.

dx/dt = 40 - 10t ⇒ x = ∫(40 - 10t) dt = 40t - 5t² + C.

Given x(0) = 0 ⇒ C = 0, so x = 40t - 5t².

Set x = 60:

60 = 40t - 5t² ⇒ t² - 8t + 12 = 0 ⇒ (t - 2)(t - 6) = 0.

Hence t = 2 s or t = 6 s. (These two times correspond to the two instants when the particle is at x = 60 m.)

Ques 7: A particle moves rectilinearly with initial velocity u and a constant acceleration a. Find the average velocity of the particle in a time interval from t = 0 to t = t second of its motion.

Ans:

Sol: 

Displacement in time t is s = ut + ½at². The average velocity over the interval is v_avg = s/t = u + ½at. Using v = u + at, this becomes v_avg = (u + v)/2, i.e., the arithmetic mean of initial and final velocities for constant acceleration.

Ques 8: A particle moves in a straight line with uniform acceleration. Its velocity at time t = 0 is v₁ and at time t = t is v₂. The average velocity of the particle in this time interval is 

 Is this statement true or false?

Ans: True

Sol: v₂ = v₁ + at ⇒ at = v₂ - v₁.

Average velocity = (v₁ + v₂)/2 for uniform acceleration, so the statement is true.

Ques 9: Find the average velocity of a particle released from rest from a height of 125 m over a time interval till it strikes the ground, g = 10 m/s².

Ans: 25 m/s (downwards)

Sol: For free fall from height h = 125 m starting from rest, time to reach ground is t = √(2h/g) = √(250/10) = √25 = 5 s.

Average velocity = total displacement / time = 125 m / 5 s = 25 m/s downward.

Ques 10: Velocity of a particle moving along x-axis varies with time as, v = (10 + 5t - t²) At time t = 0, x = 0. 

Find 

(a) acceleration of particle at t = 2 s 

(b) x-coordinate of particle at t = 3 s

Ans: (a) 1 m/s² 

(b) 43 .5 m

Sol: v(t) = 10 + 5t - t² ... (i).

Acceleration a(t) = dv/dt = 5 - 2t. At t = 2 s, a = 5 - 4 = 1 m/s².

To find x(t), integrate v(t): x = ∫(10 + 5t - t²) dt = 10t + (5/2)t² - (1/3)t³ + C. Given x(0) = 0 ⇒ C = 0.

At t = 3 s, x = 10×3 + (5/2)×9 - (1/3)×27 = 30 + 22.5 - 9 = 43.5 m.

Ques 11: Velocity of a particle at time t = 0 is 2 m/s. A constant acceleration of 2 m/s² acts on the particle for 2 seconds at an angle of 60° with its initial velocity. Find the magnitude of velocity and displacement of particle at the end of t = 2 s.

Ans:

Sol: 

Take the initial velocity along the x-axis: v_0x = 2 m/s, v_0y = 0. Acceleration components are a_x = 2 cos60° = 1 m/s², a_y = 2 sin60° = √3 m/s².

Change in velocity over 2 s: Δv_x = a_x·2 = 2 m/s, Δv_y = a_y·2 = 2√3 m/s.

Final velocity components: v_x = 2 + 2 = 4 m/s, v_y = 0 + 2√3 m/s. Magnitude of final velocity = √(4² + (2√3)²) = √(16 + 12) = √28 = 2√7 m/s.

Displacement in 2 s: s_x = v_0x·2 + ½ a_x·(2)² = 4 + 2 = 6 m; s_y = 0 + ½ a_y·4 = 2√3 m. Magnitude of displacement = √(6² + (2√3)²) = √(36 + 12) = √48 = 4√3 m.

∴ 

= 4√3 m

Ques 12: Velocity of a particle at any time t is 

Find acceleration and displacement of particle at t = 1 s. Can we apply 

 or not?

Ans: 

Sol: Part I

 ...(i)

Differentiate the given v(t) to find acceleration a(t) = dv/dt and evaluate at t = 1 s:

∴

Integrate v(t) to obtain displacement s(t) and apply the initial condition (usually s(0) = 0) to fix the constant:

Taking initial displacement to be zero:

Part II  

Yes. The kinematic relation s = v₀t + ½at² (or its equivalent forms) can be applied only when acceleration is constant. From the expressions above:

 if dv/dt is constant then acceleration is constant, implying initial velocity is

 and acceleration is

.

Thus, if a(t) is a constant function (as determined by differentiation), the standard constant-acceleration formulae are valid; otherwise they are not.

∴  

Ques 13: The coordinates of a particle moving in x-y plane at any time t are (2t, t²).. 

Find: 

(a) the trajectory of the particle, 
(b) velocity of particle at time t and 
(c) acceleration of particle at any time t.

Ans: (a) x² = 4y

Sol: x = 2t and y = t².

Eliminate t: t = x/2 ⇒ y = (x/2)² ⇒ x² = 4y. This is the trajectory (a parabola).

Velocity components: v_x = dx/dt = 2, v_y = dy/dt = 2t. So velocity vector v = (2, 2t).

Acceleration: a_x = d²x/dt² = 0, a_y = d²y/dt² = 2. Thus acceleration vector a = (0, 2).

or, x² = 4y

(The above is the equation to trajectory) x = 2t

∴ 

y = t²

∴ 

Thus,

Introductory Exercise 3.3

Ques 1: Figure shows the displacement-time graph of a particle moving in a straight line. Find the signs of velocity and acceleration of particle at time t = t₁ and t = t₂.

Ans: v_t1, a_t1 and a_t2 are positive while v_t2 is negative

Sol: At t = t₁ the slope of the s-t curve (which equals velocity) is positive, so v_t1 > 0. The curvature at t₁ shows the slope of the v-t curve is positive, so a_t1 > 0.

v = tan θ

The corresponding v-t graph is upward at that instant:

Acceleration at t = t₁ :

Since the s-t curve is concave upward at t₁ (angle α < 90°), acceleration a_t1 is positive and approximately constant in that region.

At t = t₂ the slope of the s-t curve is negative (so v_t2 < 0), but if the curve is still concave upward the rate of change of velocity is positive, hence a_t2 > 0.

Ques 2: A particle of mass m is released from a certain height h with zero initial velocity. It strikes the ground elastically (direction of its velocity is reversed but magnitude remains the same). Plot the graph between its kinetic energy and time till it returns to its initial position.

Ans:

Sol: Let the particle reach the ground at time t = t₁. While falling, its speed increases as v = gt (from rest), so kinetic energy KE = ½mv² = ½m(g t)² ∝ t². At the instant of elastic collision the direction of velocity reverses but the magnitude (hence KE) remains the same. During the rise after collision KE again decreases as v reduces, finally becoming zero when the particle returns to its initial height at time 2t₁. The KE vs t graph is a parabola (KE ∝ t²) up to t₁, an instantaneous jump-free reversal at t₁ (same KE), followed by the symmetric decrease to zero at 2t₁.

 i.e., KE ∝ t². While going up the velocity becomes negative (opposite direction) but KE remains positive; KE reduces to zero at time 2 t when the particle reaches its initial position.

Ques 3: A ball is dropped from a height of 80 m on a floor. At each collision, the ball loses half of its speed. Plot the speed-time graph and velocity-time graph of its motion till two collisions with the floor. [Take g = 10 m/s²]

Ans: 

Sol: Speed just before first collision: v = √(2gh) = √(2×10×80) = √1600 = 40 m/s. Time to reach ground: t = √(2h/g) = √(160/10) = 4 s.

At first collision speed becomes half: 40/2 = 20 m/s upwards. Time to reach maximum height after the first collision = v/g = 20/10 = 2 s. Time to return to the floor = 2 s. Thus the sequence of events and corresponding speeds and times are:

- 0 to 4 s: speed increases linearly from 0 to 40 m/s (free fall).

- At 4 s: instantaneous reduction to 20 m/s (after collision).

- 4 to 6 s: speed decreases linearly to 0 at the top of the rebound (2 s interval).

- 6 to 8 s: speed increases linearly back to 20 m/s on the way down and at 8 s collides again (second collision).

Plotting speed vs time gives piecewise linear segments as above (always non-negative). The velocity vs time graph is similar in magnitude but shows sign change at the collision (velocity becomes positive upwards after bounce, negative while falling).

Speed of ball (just after first collision with floor) = 20 m/s.

Time to attain maximum height after first collision = 2 s.

∴ Time for the return journey to floor = 2 s.

Corresponding velocity-time will be

Ques 4: Figure shows the acceleration-time graph of a particle moving along a straight line. After what time the particle acquires its initial velocity?

Ans: (2 + √3) s

Sol:

Let the areas under the a-t graph to the right of t = 2 cancel the positive area to the left so that net change in velocity from t = 0 is zero; the particle will then have its initial velocity. Setting the algebraic sum of areas to zero leads to

⇒ h = 2 (t - 2)

Net change in velocity = area under the a-t graph = 0 ⇒

or 3 - (t - 2)² = 0 ⇒ (t - 2)² = 3 ⇒ t - 2 = ±√3.

Taking the physical root t > 2 gives t = 2 + √3 s as the time when the net change in velocity is zero and the particle regains its initial velocity.

Introductory Exercise 3.4

Ques 1: Two balls A and B are projected vertically upwards with different velocities. What is the relative acceleration between them?
Ans: zero
Sol: Acceleration of each ball (ignoring air resistance) is -g (downwards). Relative acceleration of A with respect to B is a_A - a_B = (-g) - (-g) = 0. Thus their relative acceleration is zero.
Ques 2: In the above problem what is the shape of the graph between distance between the balls and time before either of the two collide with ground?
Ans: straight line passing through origin
Sol: Relative velocity v_rel = v_A - v_B is constant because relative acceleration is zero. Therefore relative displacement s_rel = v_rel·t, which is a straight line through the origin when plotted against t.

or   s = (v_A - v_B) t

Ques 3: A river 400 m wide is flowing at a rate of 2.0 m/s. A boat is sailing at a velocity of 10.0 m/s with respect to the water in a direction perpendicular to the river. 

(a) Find the time taken by the boat to reach the opposite bank. 

(b) How far from the point directly opposite to the starting point does the boat reach the opposite bank?

Ans: (a) 40 s

(b) 80 m

Sol: Component of boat's velocity across the river = 10.0 m/s. Width = 400 m.

Time to cross = distance / perpendicular speed = 400 / 10 = 40 s.

While crossing, the river carries the boat downstream at 2.0 m/s. Drift = river speed × time = 2 × 40 = 80 m. Thus the boat reaches the opposite bank 80 m downstream from the point directly opposite the start.

Time to cross river

Ques 4: An aeroplane has to go from a point A to another point B, 500 km away due 30° east of north. Wind is blowing due north at a speed of 20 m/s. The air-speed of the plane is 150 m/s. 

(a) Find the direction in which the pilot should head the plane to reach the point B. 

(b) Find the time taken by the plane to go from A to B.

Ans: 

(b) 50 min

Sol: Treat the ground velocity vector (from A to B) as the resultant of the plane's air-velocity vector and the wind vector. Using vector resolution (or the sine rule in the velocity triangle shown in the figure) gives the required heading of the plane so that the resultant points 30° east of north. The calculations (omitted here because the figure gives the triangle) lead to a ground speed along AB whose magnitude gives the time:

Applying the sine rule in ΔABC and resolving components as shown yields the ground speed along AB and hence the travel time for 500 km.

Numerical evaluation gives time ≃ 2 989 s ≃ 50 min.

= 2989 s

= 50 min

Ques 5: Two particles A and B start moving simultaneously along the line joining them in the same direction with acceleration of 1 m/s² and 2 m/s² and speeds 3 m/s and 1 m/s respectively. Initially A is 10 m behind B. What is the minimum distance between them?

Ans: 8 m

Sol: Let displacement of A relative to B be s (A relative to B). Relative acceleration a_rel = a_A - a_B = 1 - 2 = -1 m/s². Relative velocity v_rel = v_A - v_B = 3 - 1 = 2 m/s. Initial relative displacement s₀ = -10 m (A is 10 m behind B).

Relative displacement as a function of time: s(t) = s₀ + v_relt + ½ a_relt² = -10 + 2t - 0.5 t².

To find minimum s, set ds/dt = 0 ⇒ v_rel + a_relt = 0 ⇒ 2 - 1·t = 0 ⇒ t = 2 s.

At t = 2 s, s = -10 + 4 - 2 = -8 m. The negative sign indicates A is still behind B by 8 m, so the minimum distance between them is 8 m.