Starting at time t=0 from the origin with spe | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the trajectory $y = \frac{x^2}{2}$.At $t=0$,the particle is at $(0, 0)$ with speed $v = 1 	ext{ m/s}$.Differentiating $y = \frac{x^2}{2}$ w

Vedclass · Accepted Answer

$A, B, C, D$

Vedclass · Answer

(B) Given the trajectory $y = \frac{x^2}{2}$.At $t=0$,the particle is at $(0, 0)$ with speed $v = 1 	ext{ m/s}$.Differentiating $y = \frac{x^2}{2}$ with respect to $t$,we get $\frac{dy}{dt} = x \frac{dx}{dt}$,which means $v_y = x v_x$.Differentiating again with respect to $t$,we get $a_y = v_x^2 + x a_x$.$(A)$ If $a_x = 1 	ext{ m/s}^2$ and the particle is at the origin $(x=0)$,then $a_y = v_x^2 + (0)(1) = v_x^2$. Since the speed is $1 	ext{ m/s}$ and at the origin $v_y = x v_x = 0$,we have $v_x = 1 	ext{ m/s}$. Thus,$a_y = 1^2 = 1 	ext{ m/s}^2$. This is correct.$(B)$ If $a_x = 0$,then $v_x$ is constant. Since $v_x(0) = 1 	ext{ m/s}$,$v_x = 1 	ext{ m/s}$ for all $t$. Then $a_y = v_x^2 + x a_x = 1^2 + x(0) = 1 	ext{ m/s}^2$ for all $t$. This is correct.$(C)$ At $t=0$,$x=0$. Since $v_y = x v_x$,we have $v_y = 0 \cdot v_x = 0$. Thus,the velocity vector is $\vec{v} = v_x \hat{i} + 0 \hat{j}$,which points in the $x$-direction. This is correct.$(D)$ If $a_x = 0$,then $v_x = 1 	ext{ m/s}$ and $a_x = 0$. From $a_y = v_x^2 + x a_x$,we get $a_y = 1^2 + 0 = 1 	ext{ m/s}^2$. Since $a_y$ is constant,$v_y = a_y t = 1 \cdot t = t$. At $t=1 	ext{ s}$,$v_y = 1 	ext{ m/s}$. The angle $	heta$ with the $x$-axis is given by $	an 	heta = \frac{v_y}{v_x} = \frac{1}{1} = 1$,so $	heta = 45^{\circ}$. This is correct.Therefore,all statements $(A), (B), (C), (D)$ are correct.

Similar Questions

The line $\frac{x}{a} + \frac{y}{b} = 1$ moves in such a way that $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{2c^2},$ where $a, b, c \in R_0$ and $c$ is a constant. Then,the locus of the foot of the perpendicular from the origin on the given line is -

If a variable line is moving such that the intercepts made by it on the coordinate axes are reciprocal to each other,then the points $P(x, y)$ on such lines satisfy

The set of all points that forms a triangle of area $15$ sq units with the points $(1, -2)$ and $(-5, 3)$ lies on

Let $S$ be the set of points whose abscissae and ordinates are natural numbers. Let $P \in S$ be such that the sum of the distances of $P$ from $(8,0)$ and $(0,12)$ is minimum among all elements in $S$. Then,the number of such points $P$ in $S$ is

If the point $\left(\alpha, \frac{7 \sqrt{3}}{3}\right)$ lies on the curve traced by the mid-points of the line segments of the lines $x \cos \theta + y \sin \theta = 7, \theta \in \left(0, \frac{\pi}{2}\right)$ between the coordinate axes,then $\alpha$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module