The displacement and the increase in the velo | vedclass

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

Question

(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 . . . . . . $m$.

Similar Questions

$A$ car moving along a straight highway with a speed of $126 \; km/h$ is brought to a stop within a distance of $200 \; m$. How long (in $seconds$) does it take for the car to stop?

$A$ body moving with uniform acceleration travels a distance of $25 \ m$ in the fourth second and $37 \ m$ in the sixth second. The distance covered by the body in the next two seconds is: (in $m$)

The deceleration experienced by a moving motor boat,after its engine is cut-off,is given by $\frac{dv}{dt} = -kv^3$,where $k$ is a constant. If $v_0$ is the magnitude of the velocity at cut-off,the magnitude of the velocity at a time $t$ after the cut-off is:

$A$ car moving along a straight line is brought to a stop within a distance of $200 \,m$ and in a time of $10 \,s$. The initial speed of the car is (in $\,ms^{-1}$)

The acceleration of an object increases with time as $a = bt$. The object starts from the origin with an initial velocity $v_0$. Find the distance traveled by the object in time $t$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module