A body travels for 15, s starting from rest w | vedclass

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

Question

(C) For a body starting from rest $(u = 0)$ with constant acceleration $(a)$,the distance traveled in time $t$ is given by $S = \frac{1}{2}at^2$.Let t

Vedclass · Accepted Answer

${S_1} = \frac{1}{3}{S_2} = \frac{1}{5}{S_3}$

Vedclass · Answer

(C) For a body starting from rest $(u = 0)$ with constant acceleration $(a)$,the distance traveled in time $t$ is given by $S = \frac{1}{2}at^2$.Let the time interval be $T = 5\, s$.Distance in the first $5\, s$ $(t=T)$: ${S_1} = \frac{1}{2}aT^2$.Distance in the first $10\, s$ $(t=2T)$: ${S_1} + {S_2} = \frac{1}{2}a(2T)^2 = 4 	imes (\frac{1}{2}aT^2) = 4{S_1}$. Thus,${S_2} = 3{S_1}$.Distance in the first $15\, s$ $(t=3T)$: ${S_1} + {S_2} + {S_3} = \frac{1}{2}a(3T)^2 = 9 	imes (\frac{1}{2}aT^2) = 9{S_1}$. Thus,${S_3} = 9{S_1} - ({S_1} + {S_2}) = 9{S_1} - 4{S_1} = 5{S_1}$.Therefore,the ratio is ${S_1}:{S_2}:{S_3} = 1:3:5$.This implies ${S_1} = \frac{1}{3}{S_2} = \frac{1}{5}{S_3}$.

$A$ body travels for $15\, s$ starting from rest with constant acceleration. If it travels distances ${S_1}$,${S_2}$,and ${S_3}$ in the first $5\, s$,the second $5\, s$,and the next $5\, s$ respectively,the relation between ${S_1}$,${S_2}$,and ${S_3}$ is:

Similar Questions

Two bodies $A$ and $B$ start from rest from the same point with a uniform acceleration of $2\,m/s^2$. If $B$ starts one second later,then the two bodies are separated,at the end of the next second,by $.............\,m$.

Derive the equations of motion for constant acceleration using the method of calculus.

$A$ particle starts moving with acceleration $2 \, m/s^2$. The distance travelled by it in the $5^{\text{th}}$ half-second is ....... $m$. (in $.25$)

The initial velocity of a body moving along a straight line is $7 \, m/s$. It has a uniform acceleration of $4 \, m/s^2$. The distance covered by the body in the $5^{th}$ second of its motion is .......... $m$.

$A$ particle is moving with speed $v = b\sqrt{x}$ along the positive $x$-axis. Calculate the speed of the particle at time $t = \tau$ (assume that the particle is at the origin at $t = 0$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module