Set 2 K

Q1 [3 Marks]: Derive the kinematic equation $s = ut + \frac{1}{2}at^2$ using the method of integration.

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

$$v = \frac{ds}{dt}$$

Step 2: Substitute the first kinematic equation for velocity under constant acceleration ($v = u + at$) into the equation.

$$u + at = \frac{ds}{dt}$$

Step 3: Separate the variables to set up the integral.

$$ds = (u + at) dt$$

Step 4: Integrate both sides. Apply the limits: at $t = 0$, displacement $s = 0$. At time $t$, displacement is $s$.

$$\int_{0}^{s} ds = \int_{0}^{t} (u + at) dt$$

Step 5: Perform the integration. Treat initial velocity $u$ and acceleration $a$ as constants.

$$[s]_{0}^{s} = \left[ ut + \frac{at^2}{2} \right]_{0}^{t}$$

Step 6: Substitute the upper and lower limits.

$$s - 0 = \left( ut + \frac{1}{2}at^2 \right) - (0 + 0)$$

$$s = ut + \frac{1}{2}at^2 \quad (\text{Proved})$$

Q2 [3 Marks]: A body travels for one-third of its total time at velocity $v_1$ and the remaining time at velocity $v_2$. Calculate its average velocity over the entire journey.

Step 1: Let the total time of the journey be $t$.

Step 2: Calculate the distance traveled in the first part of the journey ($d_1$).
Time taken = $\frac{t}{3}$, Velocity = $v_1$.

$$d_1 = v_1 \times \left(\frac{t}{3}\right) = \frac{v_1 t}{3}$$

Step 3: Calculate the distance traveled in the remaining part of the journey ($d_2$).
Remaining time = $t - \frac{t}{3} = \frac{2t}{3}$, Velocity = $v_2$.

$$d_2 = v_2 \times \left(\frac{2t}{3}\right) = \frac{2v_2 t}{3}$$

Step 4: Find the total distance covered.

$$\text{Total Distance} = d_1 + d_2 = \frac{v_1 t}{3} + \frac{2v_2 t}{3} = \frac{t(v_1 + 2v_2)}{3}$$

Step 5: Use the formula for average velocity: Total Distance / Total Time.

$$v_{avg} = \frac{\frac{t(v_1 + 2v_2)}{3}}{t}$$

The variable $t$ cancels out from the numerator and denominator.

$$v_{avg} = \frac{v_1 + 2v_2}{3}$$

Q3 [4 Marks]: The position of an object is given by $x = 4t^2 - 15t + 20$. Find the average velocity between $t=0$ and $t=3\text{ s}$, and the instantaneous velocity at $t=3\text{ s}$.

Part A: Average Velocity
Step 1: To find average velocity, we need the initial and final positions. Calculate position at $t = 0\text{ s}$.

$$x(0) = 4(0)^2 - 15(0) + 20 = 20\text{ m}$$

Step 2: Calculate the position at $t = 3\text{ s}$.

$$x(3) = 4(3)^2 - 15(3) + 20$$

$$x(3) = 4(9) - 45 + 20 = 36 - 45 + 20 = 11\text{ m}$$

Step 3: Apply the average velocity formula: $v_{avg} = \frac{\Delta x}{\Delta t}$.

$$v_{avg} = \frac{x(3) - x(0)}{3 - 0} = \frac{11 - 20}{3} = \frac{-9}{3}$$

$$v_{avg} = -3\text{ m/s}$$

Part B: Instantaneous Velocity
Step 4: Instantaneous velocity is the derivative of position with respect to time.

$$v = \frac{dx}{dt} = \frac{d}{dt}(4t^2 - 15t + 20)$$

$$v = 8t - 15$$

Step 5: Substitute $t = 3\text{ s}$ into the velocity equation.

$$v(3) = 8(3) - 15 = 24 - 15$$

$$v(3) = 9\text{ m/s}$$

Q4 [5 Marks]: The acceleration of a particle is given by $a = 3t^2 + 2t$. If it starts from rest at the origin, find its velocity and position after 2 seconds. Does this represent uniformly accelerated motion? Justify.

Part A: Finding Velocity
Step 1: Velocity is the integral of acceleration with respect to time.

$$v = \int a \, dt = \int (3t^2 + 2t) dt$$

$$v = t^3 + t^2 + C_1$$

Step 2: Use initial conditions to find the constant $C_1$. The particle starts from rest ($v=0$ at $t=0$).

$$0 = (0)^3 + (0)^2 + C_1 \Rightarrow C_1 = 0$$

$$v(t) = t^3 + t^2$$

Step 3: Calculate velocity at $t = 2\text{ s}$.

$$v(2) = (2)^3 + (2)^2 = 8 + 4$$

$$v(2) = 12\text{ m/s}$$

Part B: Finding Position
Step 4: Position is the integral of velocity with respect to time.

$$x = \int v \, dt = \int (t^3 + t^2) dt$$

$$x = \frac{t^4}{4} + \frac{t^3}{3} + C_2$$

Step 5: Use initial conditions to find the constant $C_2$. It starts at the origin ($x=0$ at $t=0$).

$$0 = \frac{0}{4} + \frac{0}{3} + C_2 \Rightarrow C_2 = 0$$

$$x(t) = \frac{t^4}{4} + \frac{t^3}{3}$$

Step 6: Calculate position at $t = 2\text{ s}$.

$$x(2) = \frac{2^4}{4} + \frac{2^3}{3} = \frac{16}{4} + \frac{8}{3}$$

$$x(2) = 4 + 2.67$$

$$x(2) = 6.67\text{ m}$$

Part C: Justification
Step 7: Look at the given acceleration equation: $a = 3t^2 + 2t$.

No, this does NOT represent uniformly accelerated motion. Uniform acceleration means acceleration is constant. Here, acceleration depends on the variable '$t$', meaning it changes as time passes.