Chapter 3: Motion in a Straight Line

V = v₁ = ?

U = 0

h = 45 m

a = g

t = t

V = u + at

v₁ = 0 + gt

v₁ = gt

Therefore, when the ball is thrown upward, v₁ = -gt

V = v₂

u = 40 m/s

a = g

t = t

V = u + at

v₂ = 40 – gt

The relative velocity of the ball in the downward direction is – 40 m/s

But when the speed increases due to acceleration, the relative speed remains 40 m/s

a) Consider the point P(x,v) at any time t on the graph such that angle ABO is θ such that

tan θ = AQ/QP = (v₀-v)/x = v₀/x₀

When the velocity decreases from v₀ to zero during the displacement, the acceleration becomes negative.

a = – tan θ = v₀/z₀ = (v₀-v)/x

v₀ – v = (v₀/z₀)x

v = v₀(1-x/x₀) is the relation between v and x.

b) a = dv/dt = (dv/dt)(dx/dx) = (dv/dx)(dx/dt)

a = -v₀/x₀

v = -v₀/x₀

a = (v₀²x/x₀²) – (v₀²/x₀)

At x = 0

a = – v₀²/x⁰

At a = 0

The points are (0, -v₀²/x₀) and B(x₀,0)

a) Velocity attained by the rain drop which is falling freely through the height h is:

v2 = u2 – 2g(-h)

As u = 0

v = √2gh = 100√2 m/s = 510 km/h

b) Diameter of the drop, d = 2r = 4 mm

Radius of the drop, r = 2 mm = 2 × 10^-3 m

Mass of the drop, m = Vρ = 3.4 × 10^-5 kg

Momentum of the rain drop, p = mv = 5 × 10^-3 kg.m/s

c) Time required to flatten the drop is the time taken by the drop to reach the ground

t = d/v = 30 ms

d) Force exerted by a rain drop is, F = change in momentum/time = p-0/t = 168 N

e) Radius of umbrella, R = 1/2 m

Area of umbrella = 0.8 m²

No.of drops striking the umbrella, n = 320

Therefore the net force exerted = 54000N

For truck, u = 20 m/s

v = 0

a = ?

t = 5s

v = u + at

a = 4 m/s²

For car, t = 3 s

u = 20 m/s

v = 0

a = a_c

v = u + at

a_c = -20/3 m/s²

Let s be the distance between the car and the truck when the truck gives the signal and t be the time taken to cover the distance.

The human response is 0.5 s and that is the time taken by the car to cover a certain distance with uniform velocity. Therefore, (t-0.5) is the retarded motion of the car.

Velocity of car after time t,

v_c = u – at

= 20 – (20/3)(t-0.5)

Velocity of truck after time t,

v_t = 20 -4t

The bump between the car and the truck is given as:

20-(20/3)(t-0.5) = 20 – 4t

t = 5/4s

Distance travelled by the truck in time t = 21.875 m

Distance travelled by the car in time t = 23.125 m

The collision between the truck and the car distance = 1.250 m

a) For maximum velocity v(t)

dv(t)/dt = 0

Substituting the value for v, we get

t = 1.5 seconds

b) For average velocity = total distance/time taken

Average velocity = 3 m

And the average velocity is maximum when time t = 2.36 sec

c) When the acceleration is maximum in a periodic motion, the time is maximum when the body returns to the mean position when v = 0

v(t) = 6t – 2t²

When t = 3 second, acceleration is maximum

d) Distance covered between 0-3 second

s = 9m

v(t) = -(t-3) (6-t)

ds/dt = (t-3)(t-6)

Integrating the equation from 3 to 6, s = -4.5m

The net distance = 9 – 4.5 = 4.5m

Height of climb in three cycle = (4.5)(3) = 13.5m

The remaining height = 20 -13.5 = 6.5 m

Therefore, no.of cycles is 20 when the height of the pole is 4.

Let the speed of ball 1 = u₁ = 2u m/s

Then the speed of ball 2 = u₂ = u m/s

The height covered by ball 1 before coming to rest = h₁

The height covered by ball 2 before coming to rest = h₂

We know that,

v² = u² + 2gh

v² = 2gh

h = v²/2g

Therefore, h₁ = 4u²/2g

h₂ = u²/2g

From question, h₁ –h₂ = 15 m

Therefore,

4u²/2g – u²/2g = 15

u = 10 m/s

h₁ = 20 m

h₂ = 5 m

And time taken is,

t₁ = 2 sec

t₂ = 1 sec