Motion in a Straight Line — NEET UG practice

1. Set up the position equation by substituting the given values into $s(t)=\alpha t^2-\beta t+\gamma$.

Since $\alpha=1\ \text{ms}^{-2}$, $\beta=6\ \text{ms}^{-1}$, $\gamma=5\ \text{m}$, we get

$s(t)=t^2-6t+5$

2. Find the velocity by differentiating position with respect to time, since velocity is the rate of change of position.

$v(t)=\frac{ds}{dt}=2t-6$

3. Find when the velocity becomes zero, because that is when the particle reverses direction, which matters for calculating total distance (average speed depends on distance, not displacement).

Setting $v=0$:

$2t-6=0 \Rightarrow t=3\ \text{s}$

4. Determine the direction of motion in each interval. For $0<t<3$, velocity is negative (particle moves backward); for $3<t<6$, velocity is positive (particle moves forward).

5. Calculate the distance covered in each interval using the area under the $v$-$t$ graph (a triangle in each interval).

For $t=0$ to $t=3$:

$d_1=\frac{1}{2}\times 3\times 6=9\ \text{m}$

For $t=3$ to $t=6$:

$d_2=\frac{1}{2}\times 3\times 6=9\ \text{m}$

6. Add both distances to get the total distance, since average speed uses total distance, not net displacement.

$d=d_1+d_2=9+9=18\ \text{m}$

7. Apply the average speed formula: average speed equals total distance divided by total time.

$v_{avg}=\frac{d}{\Delta t}=\frac{18}{6}=3\ \text{ms}^{-1}$

Hence, the answer is D. 3.

1. Understand the physical setup. A ball thrown vertically upward experiences the acceleration due to gravity acting downward throughout its flight, both while rising and while falling.

2. Determine the nature of the acceleration. Taking the upward direction as positive, gravity acts downward and is constant in magnitude and direction at all times, so

$a=-g \ (\text{constant, throughout the motion})$

3. Relate this to the slope of the $v$-$t$ graph, since acceleration is the slope of the velocity-time curve.

$\text{slope of } v\text{-}t \text{ graph} = a = -g$

This means the graph must be a single straight line with a constant negative slope — starting from a positive velocity (going up), passing through zero at the highest point, and becoming negative (coming down).

4. Compare this requirement with the given plots. Only the plot showing a continuously negative, constant slope throughout the journey correctly represents this motion.

Hence, the answer is A. C only.

1. Recall the free-fall distance formula. When an object falls freely from rest under gravity, the distance fallen after time $t$ is

$s=\frac{1}{2}gt^2$

This shows that a longer reaction time means the ruler falls for longer, so it covers a greater distance before being caught.

2. List the reaction times given.

$t_A=0.20\,\text{s},\ t_B=0.22\,\text{s},\ t_C=0.18\,\text{s},\ t_D=0.19\,\text{s},\ t_E=0.21\,\text{s}$

3. Since distance increases with time (from the formula above), rank the times from largest to smallest to get the ranking of distances.

$t_B>t_E>t_A>t_D>t_C$

4. Apply the same order to the distances fallen, because $s$ increases with $t$.

$S_B>S_E>S_A>S_D>S_C$

Hence, the answer is C. B $>$ E $>$ A $>$ D $>$ C.