Motion in a Straight Line

Concept: An object can be considered a “point object” if the distance it covers is significantly larger than its own dimensions.

(a) A railway carriage moving without jerks between two stations:
YES. The distance between two stations (typically kilometers) is much larger than the length of the carriage (meters).

(b) A monkey sitting on top of a man cycling smoothly on a circular track:
YES. The distance covered (circumference of the track) is very large compared to the size of the monkey.

(c) A spinning cricket ball that turns sharply on hitting the ground:
NO. The ball is spinning. This means different points on the ball have different velocities. Also, the turning distance is comparable to the ball’s size.

(d) A tumbling beaker that has slipped off the edge of a table:
NO. The beaker is tumbling (rotating). The height of the table is not drastically larger than the beaker size to ignore its dimensions and rotation.

(a) (A/B) lives closer to the school than (B/A):
A lives closer than B.
Looking at the y-axis (position $x$), point P (A’s home) is lower than point Q (B’s home).

(b) (A/B) starts from the school earlier than (B/A):
A starts earlier than B.
Looking at the x-axis (time $t$), A starts at $t=0$, whereas B starts at a later time.

(c) (A/B) walks faster than (B/A):
B walks faster than A.
The slope of an $x-t$ graph represents velocity ($v = \frac{dx}{dt}$). B’s graph is steeper than A’s.

(d) A and B reach home at the (same/different) time:
Different time.
Their corresponding values on the time axis when they reach home (position P and Q) are different.

(e) (A/B) overtakes (B/A) on the road (once/twice):
B overtakes A once.
The graphs intersect at one point. Since B started later and moved faster, B crossed A on the way.

Given Data:
Distance to office $s = 2.5 \text{ km}$.
Speed to office $v_1 = 5 \text{ km h}^{-1}$.
Speed returning $v_2 = 25 \text{ km h}^{-1}$.

1. Time to reach office:

She starts at 9:00 AM and reaches at 9:30 AM.

2. Stay at office:
She stays from 9:30 AM to 5:00 PM. The graph will be a horizontal line at $x = 2.5 \text{ km}$.

3. Time to return home:

She leaves at 5:00 PM and reaches home at 5:06 PM.

Graph Summary:
– Origin (9:00 am, 0 km) to (9:30 am, 2.5 km) [Straight Line]
– (9:30 am, 2.5 km) to (5:00 pm, 2.5 km) [Horizontal Line]
– (5:00 pm, 2.5 km) to (5:06 pm, 0 km) [Steep Straight Line Down]

Motion Analysis:
Forward: 5 steps (5 m) in 5 s.
Backward: 3 steps (3 m) in 3 s.

Step-by-Step Calculation:

• Cycle 1: Time = 8s, Net Distance = 2m.
• Cycle 2: Time = 16s, Net Distance = 4m.
• Cycle 3: Time = 24s, Net Distance = 6m.
• Cycle 4: Time = 32s, Net Distance = 8m.

The Final Steps:
He is now at $8 \text{ m}$. The pit is at $13 \text{ m}$.
Remaining distance = $13 – 8 = 5 \text{ m}$.
In the next forward surge, he takes 5 steps forward. Since this takes 5 seconds, he covers exactly 5m and falls into the pit before he can step back.

Total Time:

Given:
$u = 126 \text{ km h}^{-1} = 126 \times \frac{5}{18} = 35 \text{ m s}^{-1}$.
$v = 0 \text{ m s}^{-1}$ (Stopped).
$s = 200 \text{ m}$.

(a) Retardation ($a$):
Using the third equation of motion:

Retardation magnitude is $3.06 \text{ m s}^{-2}$.

(b) Time taken ($t$):
Using the first equation of motion:

(a) Direction of acceleration during upward motion:
Acceleration due to gravity ($g$) always acts Vertically Downwards.

(b) Velocity and acceleration at highest point:
At the highest point, the ball momentarily comes to rest.
Velocity $v = 0 \text{ m s}^{-1}$.
Acceleration $a = 9.8 \text{ m s}^{-2}$ (Acting downwards).

(c) Sign Convention:
The question defines Downwards as Positive (+) x-axis direction.
Origin is at highest point ($x=0$).

• Upward Motion: Ball is below highest point moving up. Position $x > 0$. Velocity is Upward (against +ve axis) $\rightarrow v < 0$. Acceleration is Downward $\rightarrow a > 0$.
• Downward Motion: Position $x > 0$. Velocity is Downward $\rightarrow v > 0$. Acceleration is Downward $\rightarrow a > 0$.

(d) Max Height and Time:
Using $v^2 – u^2 = 2gH$ with standard convention (Up +, Down -) for calculation ease:

Time to reach top: $0 = 29.4 – 9.8t \Rightarrow t = 3 \text{ s}$.
Total time (Up + Down) = $3 + 3 = 6 \text{ s}$.

(a) With zero speed at an instant may have non-zero acceleration:
TRUE. Example: A ball thrown upwards at its highest point has $v=0$ but $a=g=9.8 \text{ m s}^{-2}$.

(b) With zero speed may have non-zero velocity:
FALSE. Speed is the magnitude of velocity. If magnitude is zero, the vector itself is zero.

(c) With constant speed must have zero acceleration:
TRUE (for 1D motion). In 1D, direction can only change by stopping (speed becomes 0). If speed is constant, velocity is constant. (Note: False for 2D/3D motion like circular motion).

(d) With positive value of acceleration must be speeding up:
FALSE. If velocity is negative and acceleration is positive, the object is slowing down. Acceleration only means force is applied in +ve direction.

Calculations:
1. Time to fall 90m: $s = ut + \frac{1}{2}gt^2 \Rightarrow 90 = 0 + 4.9t^2 \Rightarrow t \approx 4.29 \text{ s}$.
2. Velocity before impact: $v = u + gt = 0 + 9.8(4.29) \approx 42 \text{ m s}^{-1}$.
3. Rebound Velocity: Loses 1/10th speed. $v’ = 0.9v \approx 37.8 \text{ m s}^{-1}$.
4. Time to rise: $t’ = v’/g = 37.8/9.8 \approx 3.86 \text{ s}$.

Graph Nature:
It will be a “sawtooth” wave. Speed increases linearly to 42, drops instantly to 37.8, decreases linearly to 0, increases again.

Scenario:
Home $\to$ Market (2.5 km) at 5 km/h. Time $t_1 = 2.5/5 = 0.5 \text{ h} = 30 \text{ min}$.
Market $\to$ Home (2.5 km) at 7.5 km/h. Time $t_2 = 2.5/7.5 = 1/3 \text{ h} = 20 \text{ min}$.

(i) 0 to 30 min: Man moves Home to Market.
Displacement = 2.5 km. Path = 2.5 km.
Avg Velocity = $2.5/0.5 = 5 \text{ km h}^{-1}$.
Avg Speed = $5 \text{ km h}^{-1}$.

(ii) 0 to 50 min: Total trip (Home $\to$ Market $\to$ Home).
Displacement = 0. Path = 5 km.
Avg Velocity = $0 / (50/60) = \mathbf{0}$.
Avg Speed = $5 / (50/60) = 6 \text{ km h}^{-1}$.

(iii) 0 to 40 min: Home $\to$ Market (30 min) + Return journey (10 min).
Return distance in 10 min = $7.5 \times (10/60) = 1.25 \text{ km}$.
Position = $2.5 – 1.25 = 1.25 \text{ km}$ from home.
Total Path = $2.5 + 1.25 = 3.75 \text{ km}$.
Avg Velocity = $1.25 / (40/60) = 1.875 \text{ km h}^{-1}$.
Avg Speed = $3.75 / (40/60) = 5.625 \text{ km h}^{-1}$.

(a) Impossible: A vertical line drawn cuts the graph at two points. A particle cannot be at two positions at the same time.

(b) Impossible: A circle implies that at the same time instant, the particle has velocities in positive and negative directions. Also implies time flowing backwards.

(c) Impossible: Speed can never be negative. The graph shows negative speed.

(d) Impossible: Total path length can never decrease with time. The graph shows path length going down.

Given:
Police Van $v_p = 30 \text{ km h}^{-1} = \frac{25}{3} \text{ m s}^{-1}$.
Thief Car $v_t = 192 \text{ km h}^{-1} = \frac{160}{3} \text{ m s}^{-1}$.
Muzzle Speed $v_b = 150 \text{ m s}^{-1}$.

Effective Bullet Speed:
Since bullet is fired from moving van: $v_{eff} = v_b + v_p = 150 + \frac{25}{3} = \frac{475}{3} \text{ m s}^{-1}$.

Relative Impact Speed:
Speed of bullet w.r.t thief = $v_{eff} – v_t$.

(a) x-t graph (Ball Rebounding): Shows a ball striking a wall/floor and rebounding. The slope changes sign instantly.

(b) v-t graph (Free fall with gravity change?): Velocity increases, then direction flips and decreases. Could represent a ball thrown up and caught, but the oscillation suggests repeating motion.

(c) a-t graph (Impulse): Acceleration is zero, then a sharp spike, then zero. Example: A cricket ball moving at constant speed being hit by a bat for a very short duration.