Kinematics-1D

Distance covered is $x|_{t\to\infty}-x|_{t=0}=x_{0}$
Distance covered is $x|_{t\to \infty} - x|_{t = 0} = x_0$ However, since the time of motion is infinite, so average speed is zero, because
$\text { Average Speed } = \frac {\text { Total Distance Travelled }}{\text { Total Time Taken }}$
$\Rightarrow \mathrm{AverageSpeed} = \frac {x _ {0}}{\infty} = 0$
Hence, (B) and (C) are correct.

${\frac {d v}{d t}} = - v ^ {2} + 2 v - 1$
Terminal velocity is a constant velocity and hence when terminal velocity is attained, then acceleration is zero.
$\begin{array}{r l} & {\Rightarrow \frac {d v}{d t} = 0} \\ & {\Rightarrow - v ^ {2} + 2 v - 1 = 0} \\ & {\Rightarrow (v - 1) ^ {2} = 0} \\ & {\Rightarrow v = 1 \mathrm{ms} ^ {- 1}} \\ & {\mathrm{Also}, \frac {d v}{d t} = - v ^ {2} + 2 v - 1 = - (v - 1) ^ {2}} \\ & {\Rightarrow \int_ {0} ^ {v} \frac {d v}{(v - 1) ^ {2}} = - \int_ {0} ^ {t} d t} \\ & {\Rightarrow - \bigg (\frac {1}{v - 1} \bigg) \bigg | _ {0} ^ {v} = - t} \\ & {\Rightarrow \frac {1}{v - 1} + 1 = t} \\ & {\Rightarrow \frac {1}{v - 1} = t - 1} \\ & {\Rightarrow v = \frac {t}{t - 1}} \end{array}$
When acceleration is one fourth of its initial value, i.e.
$\begin{array}{r l} & a = - \frac {1}{4}, \text {then we have} \\ & \qquad - v ^ {2} + 2 v - 1 = - \frac {1}{4} \\ & \Rightarrow \quad - 4 v ^ {2} + 8 v - 3 = 0 \\ & \Rightarrow \quad v = \frac {- (8) \pm \sqrt {6 4 - 4 (- 4) (- 3)}}{2 (- 4)} \\ & \Rightarrow \quad v = \frac {- 8 \pm 4}{- 8} = \frac {1}{2} \mathrm{ms} ^ {- 1}, \frac {3}{2} \mathrm{ms} ^ {- 1} \end{array}$
Hence, (A), (B) and (D) are correct.

In x-t graph at maxima and minima. $\frac{dx}{dt}=0$ . Hence particle comes to rest for 6 times and average velocity for total period is negative.
Hence, (A) and (D) are correct.

Since, $v \frac{dv}{dx} = -\alpha v$ $\Rightarrow \int_{v_{0}}^{0} dv = -\alpha \int_{0}^{x_{0}} dx$ $\Rightarrow x_{0} = \frac{v_{0}}{\alpha}$ So, option (A) is correct.
Also, $\frac{dv}{dt} = -\alpha v$ $\Rightarrow \int_{v_{0}}^{v} \frac{dv}{v} = -\alpha \int_{0}^{t} dt$ $\Rightarrow v = v_{0} e^{-\alpha t}$ $\Rightarrow v = 0 \text{ for } t \to \infty$ So, option (C) is also correct.
Hence, (A) and (C) are correct.

Equation of straight line is $v^{2}=15x+25$ Differentiating with respect to x, we get $2v\frac{dv}{dx}=15$ $\Rightarrow v\frac{dv}{dx}=a=\frac{15}{2}\ ms^{-2}$ $\Rightarrow a=7.5\ ms^{-2}$ Since, $v=u+at=5+(7.5)(1)=12.5\ ms^{-1}$ $\Rightarrow v=12.5\ ms^{-1}$ Hence, (B), (C) and (D) are correct.

A,B,D

A,C

During accelerated motion $v = \alpha t_{1}$ and $v^{2} = 2\alpha x$ During decelerated motion $0 = v - \beta t_{2}$ $\Rightarrow \quad \alpha t_{1} = \beta t_{2}$ $\Rightarrow \quad \frac{t_{1}}{t_{2}} = \frac{\beta}{\alpha}$ Also, $0^{2} = v^{2} - 2\beta y$ $\Rightarrow \quad v^{2} = 2\beta y$ $\Rightarrow \quad 2\alpha x = 2\beta y$ $\Rightarrow \quad \frac{x}{y} = \frac{\beta}{\alpha}$
Hence, (A), (B) and (C) are correct.

Acceleration is the slope of v-t graph, which is negative between 5 s and 6 s.
Between t=0 and t=4 s the particle changes its direction of motion at 1 s and 3 s.
Hence, (C) and (D) are correct.

B,C,D

$v_{\max}=\left(\frac{a_{1}a_{2}}{a_{1}+a_{2}}\right)t=\left(\frac{2\times4}{2+4}\right)(6)=8\ \mathrm{ms}^{-1}$ and $s=\frac{1}{2}\left(\frac{a_{1}a_{2}}{a_{1}+a_{2}}\right)t^{2}=\frac{1}{2}\left(\frac{2\times4}{2+4}\right)(6)^{2}=24\ \mathrm{m}$ Hence, (A) and (C) are correct.

$v = 4t - t^{2}$ $\Rightarrow \frac{ds}{dt} = 4t - t^{2}$ $\Rightarrow \int ds = \int_{0}^{5}(4t - t^{2})dt$ $\Rightarrow s = 2 \times 5^{2} - \frac{5^{3}}{3}$ $\Rightarrow s = \frac{150 - 125}{3} = \frac{25}{3} \, m$ Average velocity $= \frac{25}{3 \times 5} \, ms^{-1} = \frac{5}{3} \, ms^{-1}$ $a = \frac{dv}{dt} = (4 - 2t) \, ms^{-2}$ $\Rightarrow a = 4 \, ms^{-2}$ at t = 0
Hence, (C) and (D) are correct.

Positive slope indicates that acceleration is increasing uniformly with time and is not uniform.
Hence, the correct answer is (D).

Acceleration is the rate of change of velocity i.e., $a = \frac{dv}{dt}$ . Hence, the correct answer is (D).

A body has no relative motion with respect to itself. Hence if a frame of reference of the body is fixed, then the bodies will be always at relatively rest in this frame of reference. So, (A) is correct
Hence, the correct answer is (A).

As $\vec{v} = \frac{d\vec{r}}{dt}$ , hence statement-I is correct.
In most of the cases known instantaneous velocity and average velocity are not equal. They can be equal only when the particle moves with a constant uniform velocity. Hence, the correct answer is (C).

To an observer on the balloon, the ground is a moving frame.
Hence, the correct answer is (D).

Time to cross the river $t = \frac{d}{\sqrt{v^{2} - u^{2}}}$

Hence, the correct answer is (A).

D

Suppose motion of 1 is viewed from 2 then as 2 is stationary. If 1 is to meet it, then 1 must come straight towards 2 i.e., the relative velocity must be along the line joining 1 and 2.
Hence, the correct answer is (A).

$\begin{array}{l l} & a _ {r} = g - g = 0 \\ & \Rightarrow \frac {d v _ {r}}{d t} = 0 \\ & \Rightarrow v _ {r} = \text { constant (say } k _ {1} \text {) } \\ & \Rightarrow \frac {d x _ {r}}{d t} = k _ {1} \\ & \Rightarrow x _ {r} = k _ {1} t \end{array}$
Hence, the correct answer is (A).

D

At the highest point of the graph velocity (slope of x-t curve) changes suddenly from a positive to a negative value. This is physically impossible.
Hence, the correct answer is (D).

Area under acceleration - time graph gives change in velocity and not average velocity.
Hence, the correct answer is (C).

The correct answer is (B).