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

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

Question

(C) Given: Initial velocity $\vec{u} = 5 \hat{j} \, m/s$,acceleration $\vec{a} = 10 \hat{i} + 4 \hat{j} \, m/s^2$,and initial position $(x_0, y_0) = (

Vedclass · Accepted Answer

$2 \, s$ and $18 \, m$

Vedclass · Answer

(C) Given: Initial velocity $\vec{u} = 5 \hat{j} \, m/s$,acceleration $\vec{a} = 10 \hat{i} + 4 \hat{j} \, m/s^2$,and initial position $(x_0, y_0) = (0, 0)$.The position vector at time $t$ is given by $\vec{r} = \vec{u}t + \frac{1}{2} \vec{a}t^2$.For the $x$-coordinate:$x = u_x t + \frac{1}{2} a_x t^2$$20 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2$$20 = 5t^2 \implies t^2 = 4 \implies t = 2 \, s$.For the $y$-coordinate:$y = u_y t + \frac{1}{2} a_y t^2$$y_0 = 5 \cdot t + \frac{1}{2} \cdot 4 \cdot t^2$Substituting $t = 2 \, s$:$y_0 = 5(2) + 2(2^2) = 10 + 8 = 18 \, m$.Thus,$t = 2 \, s$ and $y_0 = 18 \, m$.

Starting from the origin at time $t=0,$ with initial velocity $5 \hat{j} \, m/s,$ a particle moves in the $x-y$ plane with a constant acceleration of $(10 \hat{i} + 4 \hat{j}) \, m/s^2$. At time $t$,its coordinates are $(20 \, m, y_0 \, m)$. The values of $t$ and $y_0$ are respectively:

Similar Questions

When a body moves with a constant speed along a circle,

The relative angular speed of the hour hand and the minute hand of a clock is:

The kinetic energy $K$ of a particle moving along a circle of radius $R$ depends on the distance covered $s$ as $K = as^2$,where $a$ is a constant. The force acting on the particle is:

The coordinates of a moving particle at any time $t$ are given by $x = \alpha t^3$ and $y = \beta t^3$,where $\alpha$ and $\beta$ are constants. The speed of the particle at time $t$ is given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module