A body moves on a frictionless plane starting | vedclass

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

Question

(B) The distance traveled by an object in the $n^{th}$ second is given by the formula $S_n = u + \frac{a}{2}(2n-1)$.Since the body starts from rest,$u

Vedclass · Accepted Answer

$19$

Vedclass · Answer

(B) The distance traveled by an object in the $n^{th}$ second is given by the formula $S_n = u + \frac{a}{2}(2n-1)$.Since the body starts from rest,$u = 0$,so $S_n = \frac{a}{2}(2n-1)$.For $n=10$,$S_{10} = \frac{a}{2}(2(10)-1) = \frac{19a}{2}$.For $n-1=9$,$S_{9} = \frac{a}{2}(2(9)-1) = \frac{17a}{2}$.The ratio is $\frac{S_{n-1}}{S_n} = \frac{17a/2}{19a/2} = \frac{17}{19}$.Given the ratio is $1 - \frac{2}{x}$,we have $1 - \frac{2}{x} = \frac{17}{19}$.This implies $\frac{2}{x} = 1 - \frac{17}{19} = \frac{2}{19}$.Therefore,$x = 19$.

$A$ body moves on a frictionless plane starting from rest. If $S_{n}$ is the distance moved between $t=n-1$ and $t=n$,and $S_{n-1}$ is the distance moved between $t=n-2$ and $t=n-1$,then the ratio $\frac{S_{n-1}}{S_n}$ is $\left(1-\frac{2}{x}\right)$ for $n=10$. The value of $x$ is:

Similar Questions

$A$ body of mass $10 \,kg$ has a constant velocity of $10 \,m \,s^{-1}$. $A$ constant force is applied for a duration of $4 \,s$ on the object, which gives it a speed of $2 \,m \,s^{-1}$ in the opposite direction. Find the acceleration produced in it.

$A$ particle starting from rest and moving with a uniform acceleration along a straight line covers distances $a$ and $b$ in successive intervals of $p$ and $q$ seconds. The acceleration of the particle is

If the displacement of a particle varies with time as $\sqrt{x} = t + 7$,then

The motion of a particle is described by the equation $v = at$. The distance travelled by the particle in the first $4 \ s$ is: (in $a$)

An automobile travelling with a speed of $60\,km/h$ can brake to stop within a distance of $20\,m$. If the car is going twice as fast,i.e.,$120\,km/h$,the stopping distance will be ........... $m$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module