The displacement and the increase in the velo | vedclass

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(B) Let the initial velocity at time $t$ be $u$ and acceleration be $a$.Given that the increase in velocity in $1 \ s$ is $50 \ m/s$,we have $v = u +

Vedclass · Accepted Answer

$175$

Vedclass · Answer

(B) Let the initial velocity at time $t$ be $u$ and acceleration be $a$.Given that the increase in velocity in $1 \ s$ is $50 \ m/s$,we have $v = u + a(1) = u + 50$.Thus,$a = 50 \ m/s^2$.The displacement in the interval $t$ to $t+1$ is given by $s = ut + \frac{1}{2}at^2$. For the interval of $1 \ s$ starting at $t$,the displacement is $s = u(1) + \frac{1}{2}a(1)^2 = 125$.Substituting $a = 50$,we get $u + 25 = 125$,which implies $u = 100 \ m/s$.The distance travelled in the $(t+2)^{th}$ second is given by $S_n = u + \frac{a}{2}(2n - 1)$,where $n = t+2$.However,the question asks for the distance in the $(t+2)^{th}$ second relative to the start of the motion. Since $u$ is the velocity at time $t$,the distance in the next second (the $(t+1)^{th}$ second) is $125 \ m$. The distance in the $(t+2)^{th}$ second is $S = (u+a) + \frac{a}{2} = 100 + 50 + 25 = 175 \ m$.

The displacement and the increase in the velocity of a moving particle in the time interval of $t$ to $(t+1) s$ are $125 \ m$ and $50 \ m/s$,respectively. The distance travelled by the particle in $(t+2)^{th} s$ is . . . . . . $m$.

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