NEET Motion in a Straight Line Previous Year Questions with Solutions

Given the implicit trajectory relation t = x² + x, differentiate both sides with respect to time t using the chain rule:

• • v can be written in power notation: v = (2x + 1)⁻¹.

Differentiate velocity with respect to position x:

• • Now substitute into the kinematic identity for acceleration: a = v (dv/dx).

Multiply velocity by its spatial derivative:

• • The negative sign indicates retardation.
• • Matches Option (2) perfectly.

Buses depart in either direction every T minutes at constant speed v_b. The spatial separation between any two successive buses traveling in the same direction is constant:

• • T is in minutes; v_b is in km/h.

For motion in the same direction, the relative speed is (v_b - 60) km/h and the time taken to cover distance d is 30 minutes. For motion in the opposite direction, relative speed is (v_b + 60) km/h and time is 10 minutes:

• • 3(v_b - 60) = v_b + 60 ⟹ 3v_b - 180 = v_b + 60 ⟹ 2v_b = 240 ⟹ v_b = 120 km/h.

Substitute v_b = 120 km/h back into either distance expression:

• • Bus speed = 120 km/h, Bus interval T = 15 min.
• • Corresponds uniquely to Option (4).

The instantaneous acceleration a(t) is the first time derivative of velocity, which corresponds graphically to the slope of the tangent to the velocity-time graph at that instant:

Analyzing the given trapezoidal v-t curve: 1. $0 \le t < t_1$: $v = 0 \implies a = 0$. 2. $t_1 \le t < t_2$: $v$ is an upward inclined straight line with constant positive slope: $$a = \frac{dv}{dt} = +a_0 = \text{constant} > 0$$ 3. $t_2 \le t < t_3$: $v$ is a horizontal plateau with zero slope: $$a = \frac{dv}{dt} = 0$$ 4. $t_3 \le t < t_4$: $v$ is a downward inclined straight line with constant negative slope: $$a = \frac{dv}{dt} = -a_0 = \text{constant} < 0$$ 5. $t \ge t_4$: $v = 0 \implies a = 0$.

The corresponding a-t graph consists of a positive rectangular pulse, followed by a flat zero line, followed by an equal negative rectangular pulse below the time axis. This uniquely matches Graph (1).

Average speed is fundamentally defined as total distance travelled divided by total time elapsed:

Let the total distance be 2d. Each half of the journey has distance d:

Add the two time intervals together using a common denominator:

Divide total distance 2d by total time:

Assuming a uniform resisting force and therefore a constant deceleration a throughout the wooden block, use the third equation of motion v² = u² - 2as for the first stage where the bullet enters with velocity u, travels s₁ = 24 cm, and reduces its velocity to v₁ = u/3:

Rearranging the equation to find the retarding acceleration term 2a:

For the bullet to come to complete rest (v = 0) from its initial velocity u over the total length of the block s:

Substitute the value of 2a into the stopping distance equation:

Set the initial release point on the bridge as the coordinate origin y = 0. Choose the vertically upward direction as the positive y-axis. Downward vectors (acceleration due to gravity and downward displacement) carry a negative sign.

The net vertical position (displacement) y(t) of a particle moving under constant gravitational acceleration at time t is governed by:

Substitute initial velocity u = 4 m s⁻¹, elapsed time t = 4 s, and g = 10 m s⁻²:

The displacement y = -64 m signifies that the water surface lies 64 m below the level of the bridge. Therefore, the height h of the bridge above the water surface is:

For an object moving with initial velocity $u$ and uniform acceleration $a$, the distance covered between time $t = n-1$ and $t = n$ is:

Since the block starts from rest on a smooth inclined plane of angle $\theta$, $u = 0$ and $a = g\sin\theta$ (constant):

In the interval $t = n$ to $t = n+1$, the distance travelled is $S_{n+1}$:

Dividing $S_n$ by $S_{n+1}$ cancels the common acceleration factor $a/2$:

Choose downward direction as positive: - Initial velocity: $$u = +20\text{ m/s}$$ - Final velocity upon hitting ground: $$v = +80\text{ m/s}$$ - Acceleration: $$a = +g = +10\text{ m/s}^2$$ - Displacement: $$s = h$$ (height of the tower).

Because time $$t$$ is not specified, use the work-energy / kinematic relation independent of time: $$v^2 = u^2 + 2 g h$$.

$$h = \frac{v^2 - u^2}{2g} = \frac{(80)^2 - (20)^2}{2 \times 10} = \frac{6400 - 400}{20} = \frac{6000}{20} = 300\text{ m}$$.

Matches Option 4.