Set 1 Solutions - Kinematics

Q1 [3 Marks]: Derive the kinematic equation $v^2 = u^2 + 2as$ using the method of calculus.

Step 1: Start with the basic definition of instantaneous acceleration, which is the rate of change of velocity with respect to time.

$$a = \frac{dv}{dt}$$

Step 2: Apply the chain rule of differentiation to introduce displacement ($x$) into the equation.

$$a = \frac{dv}{dx} \cdot \frac{dx}{dt}$$

Step 3: We know that the rate of change of displacement is velocity ($v = \frac{dx}{dt}$). Substitute this into the equation.

$$a = v \cdot \frac{dv}{dx}$$

Step 4: Separate the variables by moving $dx$ to the left side.

$$a \, dx = v \, dv$$

Step 5: Integrate both sides with their respective limits. Let the initial velocity be $u$ at position $x=0$, and the final velocity be $v$ at position $x=s$.

$$\int_{0}^{s} a \, dx = \int_{u}^{v} v \, dv$$

Step 6: Assuming constant acceleration ($a$), bring it out of the integral and evaluate.

$$a [x]_{0}^{s} = \left[ \frac{v^2}{2} \right]_{u}^{v}$$

$$a(s - 0) = \frac{v^2}{2} - \frac{u^2}{2}$$

Step 7: Rearrange the terms to get the final standard form.

$$2as = v^2 - u^2$$

$$v^2 = u^2 + 2as \quad (\text{Proved})$$

Q2 [3 Marks]: A car travels the first half of a distance with velocity $v_1$ and the second half with velocity $v_2$. Prove that its average velocity is $\frac{2v_1v_2}{v_1+v_2}$.

Step 1: Let the total distance traveled by the car be $d$. Therefore, the distance of each half is $\frac{d}{2}$.

Step 2: Calculate the time taken for the first half of the journey ($t_1$). Time is distance divided by velocity.

$$t_1 = \frac{d/2}{v_1} = \frac{d}{2v_1}$$

Step 3: Calculate the time taken for the second half of the journey ($t_2$).

$$t_2 = \frac{d/2}{v_2} = \frac{d}{2v_2}$$

Step 4: Find the total time ($T$) taken for the entire journey.

$$T = t_1 + t_2 = \frac{d}{2v_1} + \frac{d}{2v_2}$$

Take out $\frac{d}{2}$ as a common factor:

$$T = \frac{d}{2} \left( \frac{1}{v_1} + \frac{1}{v_2} \right) = \frac{d}{2} \left( \frac{v_1 + v_2}{v_1 v_2} \right)$$

Step 5: Use the formula for average velocity, which is total displacement divided by total time.

$$v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{d}{T}$$

Step 6: Substitute the value of total time ($T$) into the equation.

$$v_{avg} = \frac{d}{\frac{d(v_1 + v_2)}{2v_1 v_2}}$$

The variable $d$ cancels out, moving the denominator fraction to the numerator.

$$v_{avg} = \frac{2v_1 v_2}{v_1 + v_2} \quad (\text{Proved})$$

Q3 [4 Marks]: The displacement of a particle is given by $x = 2t^3 - 3t^2 + 4t$. Find the time at which its acceleration becomes zero. Also find the velocity at that instant.

Step 1: Start with the given equation for displacement.

$$x = 2t^3 - 3t^2 + 4t$$

Step 2: Differentiate displacement with respect to time to find the expression for velocity ($v$).

$$v = \frac{dx}{dt} = \frac{d}{dt}(2t^3 - 3t^2 + 4t)$$

$$v = 6t^2 - 6t + 4$$

Step 3: Differentiate velocity with respect to time to find the expression for acceleration ($a$).

$$a = \frac{dv}{dt} = \frac{d}{dt}(6t^2 - 6t + 4)$$

$$a = 12t - 6$$

Step 4: The question asks for the time when acceleration becomes zero. Set $a = 0$ and solve for $t$.

$$12t - 6 = 0 \Rightarrow 12t = 6 \Rightarrow t = \frac{6}{12}$$

$$\text{Time, } t = 0.5 \text{ s}$$

Step 5: To find the velocity at this specific instant, substitute $t = 0.5$ into the velocity equation we derived in Step 2.

$$v = 6(0.5)^2 - 6(0.5) + 4$$

$$v = 6(0.25) - 3 + 4$$

$$v = 1.5 - 3 + 4 = 2.5$$

$$\text{Velocity, } v = 2.5 \text{ m/s}$$

Q4 [5 Marks]: A particle moves along the x-axis such that its velocity varies with position as $v = \alpha \sqrt{x}$, where $\alpha$ is a positive constant. Assuming it starts from the origin, derive the expressions for its position and acceleration as functions of time.

Part A: Finding Position ($x$) as a function of time ($t$)
Step 1: Write velocity as the rate of change of position.

$$v = \frac{dx}{dt} = \alpha \sqrt{x}$$

Step 2: Separate the variables by grouping $x$ terms on the left and $t$ terms on the right.

$$\frac{dx}{\sqrt{x}} = \alpha \, dt$$

$$x^{-1/2} \, dx = \alpha \, dt$$

Step 3: Integrate both sides. The particle starts from the origin, so at $t = 0$, $x = 0$. Integrate up to time $t$ and position $x$.

$$\int_{0}^{x} x^{-1/2} \, dx = \int_{0}^{t} \alpha \, dt$$

Step 4: Perform the integration using the power rule.

$$\left[ \frac{x^{1/2}}{1/2} \right]_{0}^{x} = \alpha [t]_{0}^{t}$$

$$2\sqrt{x} = \alpha t$$

Step 5: Isolate $x$ to get position as a function of time. Square both sides.

$$\sqrt{x} = \frac{\alpha t}{2}$$

$$x(t) = \frac{\alpha^2 t^2}{4}$$

Part B: Finding Acceleration ($a$) as a function of time ($t$)
Step 6: Differentiate the position expression $x(t)$ with respect to time to get velocity $v(t)$.

$$v = \frac{dx}{dt} = \frac{d}{dt} \left( \frac{\alpha^2 t^2}{4} \right)$$

$$v = \frac{\alpha^2}{4} \cdot (2t) = \frac{\alpha^2 t}{2}$$

Step 7: Differentiate the velocity expression $v(t)$ with respect to time to get acceleration $a(t)$.

$$a = \frac{dv}{dt} = \frac{d}{dt} \left( \frac{\alpha^2 t}{2} \right)$$

$$a(t) = \frac{\alpha^2}{2} \quad (\text{Constant Acceleration})$$