Relative Motion in One Dimension

Practice Questions:

Question 1: moderate

A boat takes two hours to travel 4 km down and 4 km up the river when the water is still. How much time will the boat take to make the same trip when the river starts flowing at 2 kmph?

1. 2 hour

2. 2 hour 40 minute

3. 3 hour

4. 3 hour 40 minute

View Answer

Let the boat's speed in still water be x kmph. Given

$\frac{8}{x} = 2$

, we get x = 4 kmph.

With a 2 kmph current:

Downstream speed = 6 kmph, time = $\frac{4}{6} = \frac{2}{3}$

hours
Upstream speed = 2 kmph, time = $\frac{4}{2} = 2$

hours

Total time =

$2 \frac{2}{3}$

hours or 2 hours 40 minutes.

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Question 2: moderate

A taxi leaves the station X for station Y every 10 minutes. Simultaneously, a taxi leaves the station Y also for station X every 10 minutes. The taxis move at the same constant speed and go from X to Y or vice-versa in 2 hours. How many taxis coming from the other side will each taxi meet enroute from Y to X ?

1. 10

2. 11

3. 12

4. 23

View Answer

Each taxi takes 2 hours, and taxis leave every 10 minutes.

Taxis already on the route from X to Y:

$\frac{120}{10}$ $= 12$
Taxis that start after from X to Y:

$\frac{120}{10}$ $= 12$
Exclude the simultaneously departing taxi:

$12 + 11 = 23$

Final Answer: 23 taxis.

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Question 3: moderate

A bus starts from rest moving with an acceleration of $2\text{ m/s}^2$. A cyclist, 96 m behind the bus starts simultaneously towards the bus at a speed of $20\text{ m/s}$. After what time will bus overtake the cycle :

1. 8 s

2. 10 s

3. 12 s

4. 1 s

View Answer

Let the bus's position be $x_b = (1/2)at^2 = t^2$. The cyclist's position (starting 96m behind) is $x_c = 20t - 96$. For meeting, $t^2 = 20t - 96 \Rightarrow t^2 - 20t + 96 = 0$. Factoring gives $(t-8)(t-12) = 0$, so $t=8$s or $t=12$s. At $t=8$s, cyclist is faster than bus. At $t=12$s, bus is faster, so it overtakes.

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Question 4: moderate

A boat takes 2 hours to go 10 km and come back in still water lake. The time taken for going 10 km upstream and coming back with water velocity of $5\text{ km/h}$.

1. 140 min

2. 150 min

3. 160 min

4. 170 min

View Answer

In still water, total distance is 20 km in 2 hours, so boat speed $v_b = 10\text{ km/h}$. Water velocity $v_w = 5\text{ km/h}$. Upstream speed $v_b - v_w = 5\text{ km/h}$. Time upstream = $10/5 = 2$ hours. Downstream speed $v_b + v_w = 15\text{ km/h}$. Time downstream = $10/15 = 2/3$ hours. Total time = $2 + 2/3 = 8/3$ hours = $160$ minutes.

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Question 5: moderate

Two bodies are held separated by $9.8\text{ m}$ vertically one above the other. They are released simultaneously to fall freely under gravity. After $2\text{ s}$ the relative distance between them is :

1. 4.9 m

2. 19.6 m

3. 9.8 m

4. 39.2m

View Answer

Both bodies are released simultaneously and fall under gravity. Their acceleration is identical ($g$). Since their initial relative velocity is zero and relative acceleration is zero, their relative distance remains constant. Thus, after $2$ s, the relative distance is still $9.8$ m.

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Question 6: moderate

While sitting on a tree branch $20\text{m}$ above the ground, you drop a walnut. When the walnut has fallen $5\text{m}$ you throws a second walnut straight down. What initial speed must you give the second walnut if they are both to reach the ground at the same time? (g=$10\text{m/s}^2$)

1. 5 m$s^{-1}$

2. 10 m$s^{-1}$

3. 15 m$s^{-1}$

4. None of these

View Answer

First walnut (W1): Time to fall 5m from rest is $t_1 = \sqrt{2s/g} = \sqrt{2 \cdot 5/10} = 1$ s. Total time for W1 to reach ground from 20m is $t_{total} = \sqrt{2 \cdot 20/10} = 2$ s. Second walnut (W2) must fall 15m in $t_{W2} = t_{total} - t_1 = 2-1 = 1$ s. Using $s = v_0t + (1/2)gt^2$: $15 = v_0(1) + (1/2)(10)(1)^2 \Rightarrow 15 = v_0 + 5 \Rightarrow v_0 = 10\text{ m/s}$.

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Question 7: moderate

Two bodies separated by a distance of $‘s’$ start moving towards each other with speeds of $v$ and $2v$ respectively. The uniform acceleration with which the first body should move so that they meet at the middle is:

1. $\frac{v^2}{s}$

2. $\frac{v^2}{2s}$

3. $\frac{8v^2}{s}$

4. $\frac{4v^2}{s}$

View Answer

The second body travels $s/2$ at constant speed $2v$ in time $t = s/(4v)$. For the first body: $s/2 = vt + \frac{1}{2}at^2$. Substituting $t$ gives $s/2 = s/4 + a s^2 / (32v^2)$, which simplifies to $a = 8v^2/s$.

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Question 8: moderate

When a motorcycle moving with a uniform speed $11\text{ m/s}$ is at a distance $24\text{ m}$ from a car, the car starts from rest and moves with a uniform acceleration $2\text{ m/s}^2$ away from the motorcycle. If the car begins motion at $t = 0$, time at which the motorcycle will overtake the car is $t = $:

1. $8\text{ sec}$

2. $6\text{ sec}$

3. $3\text{ sec}$

4. $1.5\text{ sec}$

View Answer

Distance equation for meeting: $11t = 24 + \frac{1}{2}(2)t^2 \Rightarrow t^2 - 11t + 24 = 0$. Solving this quadratic equation gives $t = 3\text{ s}$ and $t = 8\text{ s}$. The first overtake occurs at $t = 3\text{ s}$.

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