Motion in 1D

Step 1: Let the total time of fall be $t$ seconds. The total height is $h = \frac{1}{2}gt^2$.
Step 2: The distance covered in the last (i.e., $t$-th) second is given by $S_t = \frac{1}{2}g(2t - 1)$.
Step 3: According to the problem, $S_t = \frac{9h}{25}$. Substituting the expressions, we get $\frac{1}{2}g(2t - 1) = \frac{9}{25} \times \frac{1}{2}gt^2$.
Step 4: Canceling $\frac{1}{2}g$ from both sides, we get $2t - 1 = \frac{9}{25}t^2$, which leads to the quadratic equation $9t^2 - 50t + 25 = 0$.
Step 5: Solving the quadratic equation by factorization, $(9t - 5)(t - 5) = 0$, giving $t = 5$ s (since $t$ must be greater than 1 for a last second to exist).
Step 6: Now, calculate the height $h$ using $t = 5$ s and $g = 9.8 \text{ ms}^{-2}$: $h = \frac{1}{2} \times 9.8 \times (5)^2 = 4.9 \times 25 = 122.5 \text{ m}$. (Note: Option B contains a typographical error in the source text as 1225 m, but it is the intended correct choice).
Step 7: Thus, the height $h$ is 122.5 m, corresponding to option B.

Step 1: The distance covered in the first second is $x = \frac{1}{2}g(1)^2 = \frac{g}{2}$.
Step 2: Let the total time taken to reach the ground be $n$ seconds. The distance covered in the last (i.e., $n$-th) second is $S_n = \frac{1}{2}g(2n - 1)$.
Step 3: According to the problem, the distance covered in the last second is $7x$. So, $\frac{1}{2}g(2n - 1) = 7 \times \frac{g}{2}$.
Step 4: Canceling $\frac{g}{2}$ from both sides, we get $2n - 1 = 7$.
Step 5: Solving for $n$, we get $2n = 8 \Rightarrow n = 4$ s.
Step 6: Therefore, the body takes 4 seconds to reach the ground.

Step 1: The given displacement equation is $x = 18t + 5t^2$.
Step 2: Differentiate displacement with respect to time to find velocity: $v = \frac{dx}{dt} = 18 + 10t$.
Step 3: Differentiate velocity with respect to time to find acceleration: $a = \frac{dv}{dt} = 10 \text{ ms}^{-2}$.
Step 4: Since the acceleration is a constant value of $10 \text{ ms}^{-2}$, it does not change over time.
Step 5: Therefore, the average acceleration during any time interval, including $t_{1} = 2\text{ s}$ to $t_{2} = 4\text{ s}$, is exactly $10 \text{ ms}^{-2}$.

Step 1: The given displacement equation is $s = 6 + 12t - 2t^2$.
Step 2: Find the velocity by differentiating displacement: $v = \frac{ds}{dt} = 12 - 4t$.
Step 3: Determine when the particle changes direction by setting velocity to zero: $12 - 4t = 0 \Rightarrow t = 3 \text{ s}$.
Step 4: Calculate the displacement at the start ($t = 0$), when it changes direction ($t = 3$), and at the end of the interval ($t = 5$).
Step 5: At $t = 0$, $s_0 = 6 \text{ m}$. At $t = 3$, $s_3 = 6 + 12(3) - 2(3)^2 = 24 \text{ m}$. At $t = 5$, $s_5 = 6 + 12(5) - 2(5)^2 = 16 \text{ m}$.
Step 6: Calculate the distance covered in the first interval ($t = 0$ to $t = 3$): $d_1 = |s_3 - s_0| = |24 - 6| = 18 \text{ m}$.
Step 7: Calculate the distance covered in the second interval ($t = 3$ to $t = 5$): $d_2 = |s_5 - s_3| = |16 - 24| = 8 \text{ m}$.
Step 8: The total distance covered in the first 5 seconds is $d_1 + d_2 = 18 + 8 = 26 \text{ m}$.

Step 1: The given displacement equation is $x = \alpha t^3 + \beta t^2 + \gamma t + \delta$.
Step 2: Differentiate displacement to find the velocity equation: $v = \frac{dx}{dt} = 3\alpha t^2 + 2\beta t + \gamma$.
Step 3: Evaluate velocity at $t = 0$ to find the initial velocity: $v_0 = \gamma$.
Step 4: Differentiate velocity to find the acceleration equation: $a = \frac{dv}{dt} = 6\alpha t + 2\beta$.
Step 5: Evaluate acceleration at $t = 0$ to find the initial acceleration: $a_0 = 2\beta$.
Step 6: Form the ratio of initial acceleration to initial velocity: $\frac{a_0}{v_0} = \frac{2\beta}{\gamma}$.
Step 7: Observe that this ratio depends only on the constants $\beta$ and $\gamma$.

Step 1: The given acceleration is $a = bt$.
Step 2: Use the relation $a = \frac{dv}{dt}$ to write $dv = bt \, dt$.
Step 3: Integrate both sides with the initial condition $v = v_0$ at $t = 0$: $\int_{v_0}^{v} dv = \int_{0}^{t} bt \, dt$.
Step 4: Evaluate the integral to find velocity: $v - v_0 = \frac{1}{2}bt^2 \Rightarrow v = v_0 + \frac{1}{2}bt^2$.
Step 5: Use the relation $v = \frac{ds}{dt}$ to write $ds = (v_0 + \frac{1}{2}bt^2) dt$.
Step 6: Integrate both sides with the initial condition $s = 0$ at $t = 0$: $\int_{0}^{s} ds = \int_{0}^{t} (v_0 + \frac{1}{2}bt^2) dt$.
Step 7: Evaluate the integral to find distance: $s = v_0 t + \frac{1}{2}b \left( \frac{t^3}{3} \right) = v_0 t + \frac{1}{6}bt^3$.

Step 1: The given acceleration is $a = 3t^2 + 2t + 2$.
Step 2: Use the relation $a = \frac{dv}{dt}$ to write $dv = (3t^2 + 2t + 2) dt$.
Step 3: Integrate both sides using the initial condition $v = 2$ at $t = 0$, and find $v$ at $t = 2$: $\int_{2}^{v} dv = \int_{0}^{2} (3t^2 + 2t + 2) dt$.
Step 4: Evaluate the definite integral on the right side: $[t^3 + t^2 + 2t]_{0}^{2} = (2^3 + 2^2 + 2(2)) - 0 = 8 + 4 + 4 = 16$.
Step 5: Equate the integrated sides: $v - 2 = 16$.
Step 6: Solve for the final velocity: $v = 16 + 2 = 18 \text{ ms}^{-1}$.

Step 1: The given displacement equation is $s = t^3 - 6t^2 + 18t + 9$.
Step 2: Differentiate displacement to find the velocity equation: $v = \frac{ds}{dt} = 3t^2 - 12t + 18$.
Step 3: To find the minimum velocity, differentiate velocity to find acceleration and set it to zero: $a = \frac{dv}{dt} = 6t - 12 = 0$.
Step 4: Solve for time $t$: $6t = 12 \Rightarrow t = 2 \text{ s}$.
Step 5: Verify it is a minimum by checking the second derivative of velocity: $\frac{d^2v}{dt^2} = 6$, which is greater than zero, confirming a minimum.
Step 6: Substitute $t = 2$ into the velocity equation to find the minimum velocity: $v_{min} = 3(2)^2 - 12(2) + 18 = 12 - 24 + 18 = 6 \text{ ms}^{-1}$.