Chapter 3: Motion in a Straight Line

The correct option is b) speed of ball changes every 10 seconds, c) average speed of ball over any 20 seconds intervals is fixed and d) the acceleration of ball is the same as from the train

a) matches with iii) has a point with zero displacement for t > 0

b) matches with ii) has x>0 throughout and has a point with v = 0 and a point with a = 0

c) matches with iv) has v < 0 and a > 0

d) matches with i) has v > 0 and a < 0 throughout

The force which is generated by the bat is known as impulsive force. When the effect of gravity is ignored, it can be said that the ball moves with a uniform speed horizontally and returns back to the bat with the same speed. When the ball hits the bat, the acceleration is zero. Acceleration is generated when the ball strikes the bat and an impulsive force is generated.

Following is the graph of variation of acceleration with respect to time:

When an equation has sine and cosine functions, the nature is periodic.

a) When the particle is moving in positive x-direction, it is given as t > sin t

When the displacement is as a function of time, it is given as x(t) = t – sin t

When the equation is differentiated with respect to time, we get

Velocity v(t) = dx(t)/dt = 1 – cos t

Differentiating the above equation again with respect to time, we get

Acceleration, a(t) = dv/ dt = sin y

When t = 0, x(t) = 0

When t = π, x(t) = π > 0

When t = 0, x(t) = 2 π > 0

b) The equation is given as

x(t) = sin t

v = (d/dt)x(t) = cos t

a = dv/dt = -sin t

At t = 0, x = 0, v = 1 and a = 0

At t = π/2, x = 1, v = 0 and a = -1

At t = π, x = 0, v = -1, and a = 0

At t = 3 π/2, x = -1, v = 0 and a = 1

Therefore, it can be said that when the particle is moving along the positive x-direction, the particle comes to rest periodically and moves backward. When the displacement and velocity is involved, that is sin t and cos t, the equations are represented periodic in nature.

Let the motion be represented as:

x(t) = A + Be^{– γ t}

Let A>B and γ >0

Velocity is x(t) = dx/dt = -Be^{– γ t}

Acceleration is a(t) = dx/dt = B γ 2e^{– γ t}

Therefore, it can be said that x(t) > 0, v(t) < 0, and a>0

The concept used in this question will be based on the behaviour of a spherical object when it is dropped through a viscous fluid. When a spherical body of radius r is dropped, it is first accelerated and gradually the acceleration comes to zero, attaining a constant velocity which is known as terminal velocity.

Given,

a = g – bv

We know that,

a = dv/dt = 0 for uniform motion

g = gravitational acceleration

Therefore, it can be said that as the speed increases, acceleration decreases. When the speed is v₀, acceleration will be zero and speed remains constant.

Therefore, a = g – bv₀ = 0

v₀ = g/b

a) At t=0 and v=0 , v-t graph is:

b) At x = 0, a-t graph is:

a) x(t) = x₀ (1 – e^-γt)

v(t) = dx(t)/dt = +x₀ γ e^-γt

a(t) = dv/dt = x₀ γ² e^-γt

v(0) = x₀ γ

b) x(t) is minimum at t = 0 since t = 0 and [x(t)]min = 0

x(t) is maximum at t = ∞ since t = ∞ and [x(t)]max = e-γt = ∞

v(t) is maximum at t = 0 since t = 0 and v(0) = x₀γ

v(t) is minimum at t = ∞ since t = ∞ and v(∞) = 0

a(t) is maximum at t = ∞ since t = ∞ and a(∞) = 0

a(t) is minimum at t = 0 since t = 0 and a(0) = -x₀ γ²

The relative speed of the cars = 27 + 18 = 45 km/h

When the two cars meet together, time t is given as

t = distance between cars/relative speed of cars

= 36/(27+18)

t = 4/5 h

Therefore, distance covered by the bird in 4/5 hours = (36)(4/5) = 28.8 km

For a free fall at 9m, the horizontal distance covered by the man should be at least 10 m.

u = 0

a = 10 m/s2

s = 9 m

t = t

s = ut + 1/2 at2

Substituting the values, we get

t = √9/3 = 3/√5 sec

The horizontal distance covered by the person,

vt = (9)( 3/√5) = 12.07 m

Therefore, 12.07 m is the distance covered during a free fall of 9 m.

Therefore, when the free fall is 10 m it is 12.07-10 = 2.07 m