The acceleration of a particle starting from | vedclass

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

Question

(A) Given: Initial velocity $u = 0$,acceleration $a = 8 	ext{ m/s}^2$.Using the equation of motion $S = ut + \frac{1}{2}at^2$,since $u = 0$,we have $

Vedclass · Accepted Answer

$\frac{\sqrt{2}-1}{2} 	ext{ s}$

Vedclass · Answer

(A) Given: Initial velocity $u = 0$,acceleration $a = 8 	ext{ m/s}^2$.Using the equation of motion $S = ut + \frac{1}{2}at^2$,since $u = 0$,we have $S = \frac{1}{2}at^2$.$1$. Time taken to cover the first metre $(S_1 = 1 	ext{ m})$:$1 = \frac{1}{2} 	imes 8 	imes t_1^2 \implies 1 = 4t_1^2 \implies t_1^2 = \frac{1}{4} \implies t_1 = \frac{1}{2} 	ext{ s}$.$2$. Time taken to cover the first two metres $(S_2 = 2 	ext{ m})$:$2 = \frac{1}{2} 	imes 8 	imes t_2^2 \implies 2 = 4t_2^2 \implies t_2^2 = \frac{1}{2} \implies t_2 = \frac{1}{\sqrt{2}} 	ext{ s}$.$3$. Time taken to move the second metre is the difference between the time taken to cover $2 	ext{ m}$ and $1 	ext{ m}$:$\Delta t = t_2 - t_1 = \frac{1}{\sqrt{2}} - \frac{1}{2} = \frac{\sqrt{2}-1}{2} 	ext{ s}$.

The acceleration of a particle starting from rest and moving in a straight line with uniform acceleration is $8 \text{ m/s}^2$. The time taken by the particle to move the second metre is

Similar Questions

The displacement $S$ of a moving particle at a time $t$ is given by $S=5+\frac{48}{t}+t^3$. Then its acceleration when the velocity is zero,is

If the error committed in measuring the radius of the circle is $0.05 \%$,then the corresponding error in calculating the area is (in $\%$)

The radius of a cylinder is increasing at a rate of $3 \text{ m/s}$ and its height is decreasing at a rate of $4 \text{ m/s}$. Find the rate of change of its volume when the radius is $4 \text{ m}$ and the height is $6 \text{ m}$.

$A$ particle moves along a curve $y = \frac{2x^3 - 1}{3}$. The points on the curve at which the $y$-coordinate is changing $18$ times as fast as the $x$-coordinate are

The displacement of a particle at the time $t$ is given by $s = \sqrt{1+t}$. Then its acceleration $a$ is proportional to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module