4 Marks Questions

Question 14 Marks

Two friends A and B are standing a distance x apart in an open field and wind is blowing from A to B. A beats a drum and B hears the sound t₁ time after he sees the event. A and B interchange their positions and the experiment is repeated. This time B hears the drum t₂ time after he sees the event. Calculate the velocity of sound in still air v and the velocity of wind u. Neglect the time light takes in travelling between the friends.

Answer

Velocity of sound v, Velocity of air u, Distance between A and B be x.

In the first case, resultant velocity of sound $\text{= v + u}$

$\Rightarrow(\text{v + u})\text{t}_1=\text{x}$

$\Rightarrow\text{v + u}=\frac{\text{x}}{\text{t}_1} \ ...(1)$

In the second case, resultant velocity of sound $=\text{v}-\text{u}$

$\therefore(\text{v}-\text{u})\text{t}_2=\text{x}$

$\Rightarrow\text{v}-\text{u}=\frac{\text{x}}{\text{t}_2} \ ...(2)$

From (1) and (2)

$2\text{v}=\frac{\text{x}}{\text{t}_1}+\frac{\text{x}}{\text{t}_2}=\text{x}\Big(\frac{1}{\text{t}_1}+\frac{1}{\text{t}_2}\Big)$

$\Rightarrow\text{v}=\frac{\text{x}}{2}\Big(\frac{1}{\text{t}_1}+\frac{1}{\text{t}_2}\Big)$

From (1) $\text{u}=\frac{\text{x}}{\text{t}_1}-\text{v}=\frac{\text{x}}{\text{t}_1}-\Big(\frac{\text{x}}{2\text{t}_1}+\frac{\text{x}}{2\text{t}_2}\Big)=\frac{\text{x}}{2}\Big(\frac{1}{\text{t}_1}-\frac{1}{\text{t}_2}\Big)$

$\therefore$ Velocity of air $\text{ V}=\frac{\text{x}}{2}\Big(\frac{1}{\text{t}_1}+\frac{1}{\text{t}_2}\Big)$

And Velocity of wind $\text{u}=\frac{\text{x}}{2}\Big(\frac{1}{\text{t}_1}-\frac{1}{\text{t}_2}\Big)$

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Question 24 Marks

An elevator is descending with uniform acceleration. To measure the acceleration, a person in the elevator drops a coin at the moment the elevator starts. The coin is 6ft above the floor of the elevator at the time it is dropped. The person observes that the coin strikes the floor in 1 second. Calculate from these data the acceleration of the elevator.

Answer

For elevator and coin u = 0

As the elevator descends downward with acceleration a' (say)

The coin has to move more distance than 1.8m to strike the floor. Time taken t = 1 sec.

$\text{S}_{\text{c}}=\text{ut}+\frac{1}{2}\text{a}'\text{t}^2=0+\frac{1}{2}\text{g}(1)^2=\frac{1}{2}\text{g}$

$\text{S}_{\text{e}}=\text{ut}+\frac{1}{2}\text{at}^2=\text{u}+\frac{1}{2}\text{a}(1)^2=\frac{1}{2}\text{a}$

Total distance covered by coin is given by $=1.8+\frac{1}{2}\text{a}=\frac{1}{2}\text{g}$

$\Rightarrow1.8+\frac{\text{a}}{2}=\frac{9.8}{2}=4.9$

$\Rightarrow\text{a}=6.2\text{m/s}^2=6.2\times3.28=20.34\text{ft/s}^2.$

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Question 34 Marks

A police jeep is chasing a culprit going on a motorbike. The motorbike crosses a turning at a speed of 72km/h. The jeep follows it at a speed of 90km/h, crossing the turning ten seconds later than the bike. Assuming that they travel at constant speeds, how far from the turning will the jeep catch up with the bike?

Answer

V_P = 90km/h = 25m/s.

V_C= 72km/h = 20m/s.

In 10 sec culprit reaches at point B from A.

Distance converted by culprit S = vt = 20 × 10 = 200m.

At time t = 10 sec the police jeep is 200m behind the culprit.

$\text{Time}=\frac{\text{s}}{\text{v}}=\frac{200}{5}=40\text{s}.$ (Relative velocity is considered).

In 40s the police jeep will move from A to a distance S, where

S = vt = 25 × 40 = 1000m = 1.0km away.

$\therefore$ The jeep will catch up with the bike, 1km far from the turning.

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Question 44 Marks

Two friends A and B are standing a distance x apart in an open field and wind is blowing from A to B. A beats a drum and B hears the sound t₁ time after he sees the event. A and B interchange their positions and the experiment is repeated. This time B hears the drum t₂ time after he sees the event. Calculate the velocity of sound in still air v and the velocity of wind u. Neglect the time light takes in travelling between the friends.

Answer

Velocity of sound v, Velocity of air u, Distance between A and B be x.

In the first case, resultant velocity of sound $\text{= v + u}$

$\Rightarrow(\text{v + u})\text{t}_1=\text{x}$

$\Rightarrow\text{v + u}=\frac{\text{x}}{\text{t}_1} \ ...(1)$

In the second case, resultant velocity of sound $=\text{v}-\text{u}$

$\therefore(\text{v}-\text{u})\text{t}_2=\text{x}$

$\Rightarrow\text{v}-\text{u}=\frac{\text{x}}{\text{t}_2} \ ...(2)$

From (1) and (2)

$2\text{v}=\frac{\text{x}}{\text{t}_1}+\frac{\text{x}}{\text{t}_2}=\text{x}\Big(\frac{1}{\text{t}_1}+\frac{1}{\text{t}_2}\Big)$

$\Rightarrow\text{v}=\frac{\text{x}}{2}\Big(\frac{1}{\text{t}_1}+\frac{1}{\text{t}_2}\Big)$

From (1) $\text{u}=\frac{\text{x}}{\text{t}_1}-\text{v}=\frac{\text{x}}{\text{t}_1}-\Big(\frac{\text{x}}{2\text{t}_1}+\frac{\text{x}}{2\text{t}_2}\Big)=\frac{\text{x}}{2}\Big(\frac{1}{\text{t}_1}-\frac{1}{\text{t}_2}\Big)$

$\therefore$ Velocity of air $\text{ V}=\frac{\text{x}}{2}\Big(\frac{1}{\text{t}_1}+\frac{1}{\text{t}_2}\Big)$

And Velocity of wind $\text{u}=\frac{\text{x}}{2}\Big(\frac{1}{\text{t}_1}-\frac{1}{\text{t}_2}\Big)$

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Question 54 Marks

A player hits a baseball at some angle. The ball goes high up in space. The player runs and catches the ball before it hits the ground. Which of the two (the player or the ball) has greater displacement?

Answer

Same displacement.

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Question 64 Marks

An elevator is descending with uniform acceleration. To measure the acceleration, a person in the elevator drops a coin at the moment the elevator starts. The coin is 6ft above the floor of the elevator at the time it is dropped. The person observes that the coin strikes the floor in 1 second. Calculate from these data the acceleration of the elevator.

Answer

For elevator and coin u = 0

As the elevator descends downward with acceleration a' (say)

The coin has to move more distance than 1.8m to strike the floor. Time taken t = 1 sec.

$\text{S}_{\text{c}}=\text{ut}+\frac{1}{2}\text{a}'\text{t}^2=0+\frac{1}{2}\text{g}(1)^2=\frac{1}{2}\text{g}$

$\text{S}_{\text{e}}=\text{ut}+\frac{1}{2}\text{at}^2=\text{u}+\frac{1}{2}\text{a}(1)^2=\frac{1}{2}\text{a}$

Total distance covered by coin is given by $=1.8+\frac{1}{2}\text{a}=\frac{1}{2}\text{g}$

$\Rightarrow1.8+\frac{\text{a}}{2}=\frac{9.8}{2}=4.9$

$\Rightarrow\text{a}=6.2\text{m/s}^2=6.2\times3.28=20.34\text{ft/s}^2.$

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Question 74 Marks

A police jeep is chasing a culprit going on a motorbike. The motorbike crosses a turning at a speed of 72km/h. The jeep follows it at a speed of 90km/h, crossing the turning ten seconds later than the bike. Assuming that they travel at constant speeds, how far from the turning will the jeep catch up with the bike?

Answer

V_P = 90km/h = 25m/s.

V_C= 72km/h = 20m/s.

In 10 sec culprit reaches at point B from A.

Distance converted by culprit S = vt = 20 × 10 = 200m.

At time t = 10 sec the police jeep is 200m behind the culprit.

$\text{Time}=\frac{\text{s}}{\text{v}}=\frac{200}{5}=40\text{s}.$ (Relative velocity is considered).

In 40s the police jeep will move from A to a distance S, where

S = vt = 25 × 40 = 1000m = 1.0km away.

$\therefore$ The jeep will catch up with the bike, 1km far from the turning.

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Question 84 Marks

A player hits a baseball at some angle. The ball goes high up in space. The player runs and catches the ball before it hits the ground. Which of the two (the player or the ball) has greater displacement?

Answer

Same displacement.

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Question 94 Marks

Complete the following table:

Car Model

Driver X
Reaction time 0.20s

Driver Y
Reaction time 0.30s

A (deceleration on hard braking = 6.0m/s²)

Speed = 54km/h
Braking distance

a = ...
Total stopping distance
b = ...

Speed = 72km/h
Braking distance

c = ...
Total stopping distance
d = ...

B (deceleration on hard braking = 7.5m/s²)

Speed = 54km/h
Breaking distance

e = ...
Total stopping distance
f = ...

Speed 72km/h
Braking distance

g = ...
Total stopping distance
h = ...

Answer

Car Model

Driver X
Reaction time 0.25s

Driver Y
Reaction time 0.35s

A (deceleration on hard braking = 6.0m/s²)

Speed = 54km/h
Braking distance

a = 19m
Total stopping distance
b = 22m

Speed = 72km/h
Braking distance

c = 33m
Total stopping distance
d = 39m

B (deceleration on hard braking = 7.5m/s²)

Speed = 54km/h
Breaking distance

e = 15m
Total stopping distance
f = 18m

Speed 72km/h
Braking distance

g = 27m
Total stopping distance
h = 33m

$\text{a}=\frac{0^2-15^2}{2(-6)}=19\text{m}$

So, b = 0.2 × 15 + 19 = 33m

Similarly other can be calculated.

Braking distance: Distance travelled when brakes are applied.

Total stopping distance = Braking distance + distance travelled in reaction time.

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Question 104 Marks

Complete the following table:

Car Model

Driver X
Reaction time 0.20s

Driver Y
Reaction time 0.30s

A (deceleration on hard braking = 6.0m/s²)

Speed = 54km/h
Braking distance

a = ...
Total stopping distance
b = ...

Speed = 72km/h
Braking distance

c = ...
Total stopping distance
d = ...

B (deceleration on hard braking = 7.5m/s²)

Speed = 54km/h
Breaking distance

e = ...
Total stopping distance
f = ...

Speed 72km/h
Braking distance

g = ...
Total stopping distance
h = ...

Answer

Car Model

Driver X
Reaction time 0.25s

Driver Y
Reaction time 0.35s

A (deceleration on hard braking = 6.0m/s²)

Speed = 54km/h
Braking distance

a = 19m
Total stopping distance
b = 22m

Speed = 72km/h
Braking distance

c = 33m
Total stopping distance
d = 39m

B (deceleration on hard braking = 7.5m/s²)

Speed = 54km/h
Breaking distance

e = 15m
Total stopping distance
f = 18m

Speed 72km/h
Braking distance

g = 27m
Total stopping distance
h = 33m

$\text{a}=\frac{0^2-15^2}{2(-6)}=19\text{m}$

So, b = 0.2 × 15 + 19 = 33m

Similarly other can be calculated.

Braking distance: Distance travelled when brakes are applied.

Total stopping distance = Braking distance + distance travelled in reaction time.

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