Kinematics-1D

The correct answer is (B).

The correct answer is (D).

The correct answer is (C). Combined solution to 1, 2, 3
For A
$\begin{array}{l} d v _ {A} = a _ {A} d t \\ \Rightarrow \int_ {0} ^ {v _ {A}} d v _ {A} = \int_ {0} ^ {t} (6 t - 3) d t \end{array}$
$\Rightarrow v _ {A} = 3 t ^ {2} - 3 t$

For B
$d v _ {B} = a _ {B} d t$
$\Rightarrow \int_ {0} ^ {v _ {B}} d v _ {B} = \int_ {0} ^ {t} (1 2 t ^ {2} - 8) d t$
$\Rightarrow v _ {B} = 4 t ^ {3} - 8 t$
Let us now calculate the times when $A$ and $B$ are at rest.
The particle $A$ is at rest $(v_{A} = 0)$ , when

The particle $B$ is at rest $(v_{B} = 0)$ , when
$\begin{array}{r l} & 4 t ^ {3} - 8 t = 0 \\ \Rightarrow & t = 0 \mathrm{s} \text {and} t = \sqrt {2} \mathrm{s} \end{array}$
The position of particles A and B can be determined using
$\begin{array}{l l} & v = \frac {d x}{d t}, \text {so} \\ & d x _ {A} = v _ {A} d t \\ \Rightarrow & \int_ {0} ^ {x _ {A}} d x _ {A} = \int_ {0} ^ {t} (3 t ^ {2} - 3 t) d t \\ \Rightarrow & x _ {A} = t ^ {3} - \frac {3}{2} t ^ {2} \end{array}$
Similarly
$\begin{array}{l} d x _ {B} = v _ {B} d t \\ \Rightarrow \int_ {0} ^ {x _ {B}} d x _ {B} = \int_ {0} ^ {t} (4 t ^ {3} - 8 t) d t \end{array}$
$\Rightarrow x _ {B} = t ^ {4} - 4 t ^ {2}$
The positions of particle A at t = 1 s and 4 s are
$x _ {A} \big | _ {t = 1 \mathrm{s}} = 1 ^ {3} - \frac {3}{2} (1 ^ {2}) = - 0. 5 \mathrm{m}$
$x _ {A} \big | _ {t = 4 \mathrm{s}} = 4 ^ {3} - \frac {3}{2} (4 ^ {2}) = 4 0 \mathrm{m}$
Particle A has travelled a total distance given by
$d _ {A} = 2 (0. 5) + 4 0 = 4 1 \mathrm{m}$
The positions of particle B at $t = \sqrt{2}$ s and 4 s are
$x _ {B} \big | _ {t = \sqrt {2}} = (\sqrt {2}) ^ {4} - 4 (\sqrt {2}) ^ {2} = - 4 \mathrm{m}$
$x _ {B} \big | _ {t = 4} = (4) ^ {4} - 4 (4) ^ {2} = 1 9 2 \mathrm{m}$
Particle B has travelled a total distance given by
$d _ {B} = 2 (4) + 1 9 2 = 2 0 0 \mathrm{m}$
At $t = 4$ s the distance between $A$ and $B$ is
$\Delta x _ {A B} = 1 9 2 - 4 0 = 1 5 2 \mathrm{m}$

Taking, $+x$ direction as positive, we get $u = 40 \text{ ms}^{-1}$ and $a = -10 \text{ ms}^{-2}$ So, this $a$ acts as retardation and the velocity of the body is first reduced to zero, say at time $t_0$ .
Since $v = u + at$ $\Rightarrow 0 = 40 + (-10)t_0$ $\Rightarrow t_0 = 4\text{ s}$ Hence, the correct answer is (B).

$\begin{array}{r l} & 0 ^ {2} - u ^ {2} = 2 a x _ {0} \\ & \Rightarrow - (4 0) ^ {2} = 2 (- 1 0) x _ {0} \\ & \Rightarrow x _ {0} = \frac {1 6 0 0}{2 0} = 8 0 \mathrm{m} \\ & \quad \longrightarrow \oplus \\ & \quad \text { } u = 4 0 \mathrm{ms} ^ {- 1} \quad x = 8 0 \mathrm{m} \\ & \quad \text { } x = 0 \quad x = 1 0 \mathrm{m} \end{array}$
So, the maximum x-coordinate of the particle, along the +x direction is
$x _ {\mathrm{MAX}} = x _ {\mathrm{Initial}} + x _ {0}$
$\Rightarrow x _ {\mathrm{MAX}} = 1 0 + 8 0 = 9 0 \mathrm{m}$
Hence, the correct answer is (C).

Further, at the origin, $s = -10 \mathrm{~m}$ . So, using $v^{2} - u^{2} = 2as$ , we get $v^{2} - (40)^{2} = 2(-10)(-10)$ $\Rightarrow v^{2} = 1800$ $\Rightarrow v = 30\sqrt{2} \mathrm{~ms}^{-1}$ , along negative $x$ direction
Hence, the correct answer is (B).

Also, using $v = u + at$ , we get $t = \frac{v - u}{a} = \frac{-30\sqrt{2} - 40}{-10}$ $\Rightarrow t = (4 + 3\sqrt{2})s$ Hence, the correct answer is (B).

$\begin{array}{l l} a = 2 t - 4 \\ & \Rightarrow \frac {d v}{d t} = 2 t - 4 \\ & \Rightarrow \int_ {0} ^ {v} d v = 2 \int_ {0} ^ {t} t d t - 4 \int_ {0} ^ {t} d t \\ & \Rightarrow v = \left(2 \frac {t ^ {2}}{2} - 4 t\right) \Bigg | _ {0} ^ {t} \\ & \Rightarrow v = t ^ {2} - 4 t \\ & \text { Now, } v = 0 \\ & \Rightarrow t ^ {2} - 4 t = 0 \\ & \Rightarrow t = 0 (\text { initially }) \text { OR } t = 4 \text { s } \\ & \text { Hence,the correct answer is(D). } \end{array}\tag{... (1}$

The velocity of the particle is maximum, when
$\begin{array}{l l} & \frac {d v}{d t} = 0 \\ \Rightarrow & a = 0 \\ \Rightarrow & 2 t _ {0} - 4 = 0 \\ \Rightarrow & t _ {0} = 2 \mathrm{s} \\ \text { and } & v _ {\max} = v | _ {t _ {0} = 2 \mathrm{s}} = 2 ^ {2} - 4 (2) \\ \Rightarrow & v _ {\max} = - 4 \mathrm{ms} ^ {- 1} \\ & \text { Hence,the correct answer is(B). } \end{array}$

Since we have calculated
$\begin{array}{r l} & v = t ^ {2} - 4 t \\ \Rightarrow & v = t ^ {2} - 4 t + 4 + 4 \\ \Rightarrow & v = (t - 2) ^ {2} + 4 \end{array}$
$\Rightarrow v - 4 = (t - 2) ^ {2}$
This happens to be the equation of the parabola with origin at the point (2, 4)
Hence, the correct answer is (D).

Initially, at t=0, the body falls from rest, so v=0 at t=0 and hence $\left.\frac{dv}{dt}\right|_{t=0}=6\ ms^{-2}$
Hence, the correct answer is (A).

Acceleration is zero (when terminal velocity is attained), so $0 = 6 - 3v_{T}$ $\Rightarrow v_{T} = 2 ms^{-1}$ Hence, the correct answer is (C).

$\frac{dv}{dt} = 6 - 3\mathrm{v}$
$\begin{array}{r l} \Rightarrow & \int_ {0} ^ {v} \frac {d v}{6 - 3 v} = \int_ {0} ^ {t} d t \\ \Rightarrow & - \frac {1}{3} \log_ {e} (6 - 3 v) \Bigg | _ {0} ^ {v} = t \\ \Rightarrow & \log_ {e} \left(\frac {6 - 3 v}{6}\right) = - 3 t \\ \Rightarrow & v = 2 (1 - e ^ {- 3 t}) \end{array}$
Hence, the correct answer is (B).

$a = \frac{dv}{dt} = 6e^{-3t}$ Hence, the correct answer is (C).

$v_{x}=a\omega\cos(\omega t)$ $v_{y}=a\omega\sin(\omega t)$ Since, $v^{2}=v_{x}^{2}+v_{y}^{2}$ $\Rightarrow \quad v = a\omega$ Hence, the correct answer is (A).

$\begin{array}x = a \sin (\omega t) = 2 a \sin \left(\frac {\omega t}{2}\right) \cos \left(\frac {\omega t}{2}\right) \\ & y = 2 a \sin^ {2} \left(\frac {\omega t}{2}\right) \\ & \Rightarrow \quad \sin \left(\frac {\omega t}{2}\right) = \sqrt {\frac {y}{2 a}} \\ & \Rightarrow \quad \cos \left(\frac {\omega t}{2}\right) = \sqrt {1 - \frac {y}{2 a}} \end{array}$
$\Rightarrow x = 2 a \sqrt {\frac {y}{2 a}} \sqrt {1 - \frac {y}{2 a}}$
$\Rightarrow x = \sqrt {2 a y} \sqrt {1 - \frac {y}{2 a}}$
Hence, the correct answer is (B).

$a _ {x} = \frac {d v _ {x}}{d t} = - a \omega^ {2} \sin (\omega t)$
$a _ {y} = \frac {d v _ {y}}{d t} = a \omega^ {2} \cos (\omega t)$
$\text { Since acceleration } = \sqrt {a _ {x ^ {2}} + a _ {y ^ {2}}}$
$\Rightarrow \quad \mathrm{Acceleration} = a \omega^ {2}$
Hence, the correct answer is (B).

Since displacement is zero, so average velocity is also zero.
Hence, the correct answer is (A).

From A to B, let time taken be $t_{1}$ , then
$t _ {1} = \frac {d}{v}$
From B to A, let time taken be $t_{2}$ , then
$t _ {2} = \frac {d}{\left(v - \frac {v}{3}\right)} = \frac {3 d}{2 v}$
So, total time taken is
$\begin{array}{r l} & T = t _ {1} + t _ {2} \\ \Rightarrow & T = \frac {d}{v} + \frac {3 d}{2 v} = \frac {5 d}{2 v} \end{array}$
Hence, the correct answer is (D).

Average speed $(v_{av}) = \frac{\text{Total Distance}}{\text{Total Time}}$ $\Rightarrow v_{av} = \frac{2d}{T}$ $\Rightarrow v_{av} = \frac{2d}{\left(\frac{5d}{2v}\right)}$ $\Rightarrow v_{av} = \frac{4v}{5}$
Hence, the correct answer is (D).

According to the problem, we have
$s _ {1} + s _ {2} = \frac {1}{2} v _ {\mathrm{max}} t$

$\Rightarrow 1 0 0 = \frac {1}{2} v _ {\max} (2 0)$
$\Rightarrow v _ {\mathrm{max}} = 1 0 \mathrm{ms} ^ {- 1}$
Hence, the correct answer is (B).

Now $v_{max} = \alpha t_{1} = \beta t_{2}$ $\Rightarrow \quad 10 = 2t_{1}$ $\Rightarrow \quad t_{1} = 5 ms^{-1}$ Now $t_{1} + t_{2} = 20 s$ $\Rightarrow \quad t_{2} = 15 s$ Hence, the correct answer is (C).

Also $v_{max} = \beta t_{2}$
$\Rightarrow 1 0 = \beta (1 5)$
$\Rightarrow \quad \beta = \frac {2}{3} \mathrm{ms} ^ {- 2}$
Hence, the correct answer is (B).

Average speed = $\frac{Total\ Distance}{Total\ Time}$ $\Rightarrow v_{av}=\frac{100}{20}=5\ ms^{-1}$ Hence, the correct answer is (A).

$s_{1}=\frac{1}{2}v_{\max}t_{1}$ $\Rightarrow \quad s_{1}=\frac{1}{2}(10)(5)=25\ m$ Hence, the correct answer is (A).