Set 3 K

Q1 [3 Marks]: Establish the relation $v = u + at$ mathematically using elementary calculus.

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

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

Step 2: Separate the variables by multiplying both sides by $dt$.

$$dv = a \, dt$$

Step 3: Integrate both sides with their proper limits. Let the initial velocity be $u$ at time $t = 0$, and the final velocity be $v$ at time $t$.

$$\int_{u}^{v} dv = \int_{0}^{t} a \, dt$$

Step 4: Assuming acceleration ($a$) is constant, bring it outside the integral sign.

$$[v]_{u}^{v} = a [t]_{0}^{t}$$

Step 5: Apply the upper and lower limits.

$$v - u = a(t - 0)$$

$$v - u = at$$

Step 6: Rearrange the equation to isolate final velocity ($v$).

$$v = u + at \quad (\text{Proved})$$

Q2 [3 Marks]: Can a body have a constant speed but a varying velocity? Can it have zero velocity but a non-zero acceleration? Give one 1D example for each.

Part A: Constant Speed but Varying Velocity
Answer: Yes. Speed is a scalar (magnitude), while velocity is a vector (magnitude + direction). If the direction changes, velocity changes even if speed is constant.
1D Example: A perfectly elastic collision of a ball against a rigid wall. Just before hitting the wall, its velocity is $+v$. Just after bouncing back, its velocity is $-v$. The speed is $|v|$ in both cases (constant), but the velocity changed from positive to negative because the direction reversed.

Part B: Zero Velocity but Non-Zero Acceleration
Answer: Yes. Acceleration is the rate of change of velocity. A body can momentarily stop (zero velocity) while its velocity is in the process of changing (non-zero acceleration).
1D Example: A stone thrown vertically upwards. At its absolute highest point (maximum height), it stops momentarily to change direction, making its instantaneous velocity exactly zero ($v = 0$). However, gravity is still acting on it, so its acceleration is $-g$ (approx $-9.8 \text{ m/s}^2$).

Q3 [4 Marks]: A particle’s velocity is given by $v = 10 + 2t^2$. Find the average acceleration of the particle between $t=2\text{ s}$ and $t=5\text{ s}$.

Step 1: Recall the formula for average acceleration. It is the total change in velocity divided by the total time interval.

$$a_{avg} = \frac{\Delta v}{\Delta t} = \frac{v(t_2) - v(t_1)}{t_2 - t_1}$$

Step 2: Calculate the initial velocity at $t_1 = 2\text{ s}$ using the given equation $v = 10 + 2t^2$.

$$v(2) = 10 + 2(2)^2$$

$$v(2) = 10 + 2(4) = 10 + 8 = 18\text{ m/s}$$

Step 3: Calculate the final velocity at $t_2 = 5\text{ s}$.

$$v(5) = 10 + 2(5)^2$$

$$v(5) = 10 + 2(25) = 10 + 50 = 60\text{ m/s}$$

Step 4: Substitute these values into the average acceleration formula.

$$a_{avg} = \frac{60 - 18}{5 - 2} = \frac{42}{3}$$

$$a_{avg} = 14\text{ m/s}^2$$

Q4 [5 Marks]: A particle moves in a straight line with an initial velocity of $5\text{ m/s}$ and a constant acceleration of $-2\text{ m/s}^2$. Find the total distance traveled by the particle in the first 4 seconds. (Hint: Watch for the turning point).

Step 1: Check for a turning point.
Because acceleration is negative (opposite to initial velocity), the particle will eventually stop and reverse direction. We must find the time when velocity becomes zero.

$$v = u + at \Rightarrow 0 = 5 + (-2)t$$

$$2t = 5 \Rightarrow t = 2.5\text{ s}$$

Since $2.5\text{ s}$ is less than the total time ($4\text{ s}$), the particle reverses direction. Distance is not equal to displacement! We must split the journey into two parts.

Step 2: Calculate distance $D_1$ (From $t = 0$ to $t = 2.5\text{ s}$)
This is the distance covered before stopping.

$$D_1 = ut + \frac{1}{2}at^2 = 5(2.5) + \frac{1}{2}(-2)(2.5)^2$$

$$D_1 = 12.5 - (6.25) = 6.25\text{ m}$$

Step 3: Calculate distance $D_2$ (From $t = 2.5\text{ s}$ to $t = 4\text{ s}$)
The new time interval for this part is $\Delta t = 4 - 2.5 = 1.5\text{ s}$.
The initial velocity for this part is $0$ (since it just stopped). The acceleration is still $-2\text{ m/s}^2$.

$$s_2 = ut + \frac{1}{2}at^2 = 0(1.5) + \frac{1}{2}(-2)(1.5)^2$$

$$s_2 = 0 - 1(2.25) = -2.25\text{ m}$$

Since distance is a scalar, we take the absolute magnitude: $D_2 = |-2.25| = 2.25\text{ m}$.

Step 4: Calculate Total Distance
Add the magnitudes of the distances from both parts of the journey.

$$D_{total} = D_1 + D_2 = 6.25 + 2.25$$

$$\text{Total Distance } = 8.5\text{ m}$$

Note for students: If you blindly used the formula $s = ut + \frac{1}{2}at^2$ for $t=4$, you would get $5(4) - 1(16) = 4\text{ m}$. That is the displacement, not the distance! Always check for turning points.