Motion in a Lane MCQs for NEET — Physics Questions with Answers

When a vector is turned through a small angle dθ without changing its length, the magnitude of the change in the vector (|Δa|) is a × dθ, where a is the original vector length. The change in the vector (Δa) itself is perpendicular to the original vector, with a magnitude of a × dθ.

Let the velocity of the scooterist =v

Relative velocity of scooterist with respect to bus = (v – 10)

$$\begin{gathered} \Rightarrow S=(v-10)\times 100 \\ \Rightarrow 1000=(v-10)\times 100 \end{gathered}$$

∴ $v=10+10=20m\text{}/\text{}s$  

(c) Given, x=5t-2$t^2$

The velocity of the particle in the x-direction,

   $v_x=\frac{dx}{dt}=\frac{d}{dt}\left(5t-2t^2\right)=5-4t$

Acceleration, $a_x=\frac{d{v}_x}{dt}=-4 m{s}^{-2}$

The velocity of the particle in the y-direction,

$$\begin{gathered} v_y=\frac{dy}{dt}=10 \\ \therefore Acceleration a_y=\frac{d{v}_y}{dt}=0 \\ \therefore Net acceleration of the particle , \\ {\overrightarrow{a}}_{net}=a_x\hat{i}+a_y\hat{j}=\left(-4m{s}^2\right)\hat{i} \end{gathered}$$

or   $a_{net}=-4 m{s}^{-2}$

(c)

$$\begin{gathered} Speed of Preeti, v_1=\frac{h}{t_1} \\ Speed of escalator, v_2=\frac{h}{t_2} \\ Speed of Preeti with escalator=v_1+v_2 \\ (Velocity of both will combine to give high velocity) \\ Time=\frac{h}{v_1+v_2}=\frac{h}{\frac{h}{t_1}+\frac{h}{t_2}}=\frac{t_1{t}_2}{t_1+t_2} \end{gathered}$$

As per the NCERT text, 'A unit vector is a vector of unit magnitude and points in a particular direction. It has no dimension and unit. It is used to specify a direction only.' If a unit vector $\hat{n}$ is multiplied by a positive scalar $\lambda$, the result is $\lambda\hat{n}$. While the magnitude changes to $\lambda$, the direction remains the same as the original unit vector. So the unit vector itself, representing direction, does not change, only the resulting vector does. Statement 4 is incorrect because the unit vector itself (its direction) does not change when multiplied by a scalar, only the magnitude of the resulting vector changes.

According to equation (3.12) in the NCERT text, 'Thus, $\vec{A} = A_x \hat{i} + A_y \hat{j}$'. This represents the resolution of a vector into its component vectors along the unit vectors $\hat{i}$ (x-axis) and $\hat{j}$ (y-axis).

From the summary point 9 in the NCERT text, 'A unit vector associated with a vector A has magnitude 1 and is along the vector A: $\hat{n} = \frac{\vec{A}}{|A|}$'.

As stated in equation (3.9) of the NCERT text, 'Since these are unit vectors, we have $|\hat{i}| = |\hat{j}| = |\hat{k}| = 1$.' A unit vector, by definition, has a magnitude of one.

The NCERT text states, 'It is convenient to resolve a general vector along the axes of a rectangular coordinate system using vectors of unit magnitude. These are called unit vectors that we discuss now. A unit vector is a vector of unit magnitude and points in a particular direction. It has no dimension and unit. It is used to specify a direction only.' Therefore, their main application is to conveniently specify direction.