A point moves linearly with deceleration which is given by `dv//dt=-alphasqrt(v)`, where alpha is a positive constant. At the start `v=v_(0)`. The distance traveled by particle before it stops will be

A point moves linearly with deceleration which is given by `dv//dt=-al

1.	A point moves linearly with deceleration which is given by `dv//dt=-alphasqrt(v)`, where alpha is a positive constant. At the start `v=v_(0)`. The distance traveled by particle before it stops will be
Answer» `(dv)/(dt)=-alphav^(1//2)` `int_(v_(0))^(v)v^(-1//2) dv=-alphaint_(0)^(t) dt` `(\|v^(-1/2+1)\|_(v_(0))^(v))/((-1/2+1))=-alpha\|t\|_(0)^(t)` `\|v^(1/2)\|_(v_(0))^(v)=alpha/2t` `sqrt(v)-sqrt(v_(0))=-alpha/2timpliesv=[sqrt(v_(0))-(alphat)/2]^(2)` The particle will stop when `v=0` `t=(2sqrt(v_(0)))/alpha` To calculate the displacement by velocity is lengthy. `(dv)/(dt)=v(dv)/(ds)=-alpha v^(1//2)` `int_(v_(0))^(v) v^(1//2) dv=-alphaint_(0)^(s) ds` `\| v^(3//2)\|_(v_(0))^(v)/(3/2)=-alphas` `v^(3//2)-v_(0)^(3//2)=-3/2alphas` When particle stops, `v=0` `impliess=(2v_(0)^(3//2))/(3alpha)`

Discussion

No Comment Found

Related InterviewSolutions

Speeds of two identical cars are u and 4u at at specific instant. The ratio of the respective distances in which the two cars are stopped from that instant isA. `1:1`B. `1:4`C. `1:8`D. `1:16`
A train starts from rest with constant acceleration `a=1 m//s^(2)` A passenger at a distance S from the train runs at this maximum velocity of `10 m//s` to catch the train at the same moment at which the train starts. Find the speed of the train when the passenger catches it for the critical distance:A. `8 m//s`B. `10 m//s`C. `12 m//s`D. `15 m//s`
A train starts from rest with constant acceleration `a=1 m//s^(2)` A passenger at a distance S from the train runs at this maximum velocity of `10 m//s` to catch the train at the same moment at which the train starts. If `S=25.5` m and passenger keeps running, find the time in which he will catch the train:A. 5 secB. 4 secC. 3 secD. `2sqrt(2)` sec
A train starts from rest with constant acceleration `a=1 m//s^(2)` A passenger at a distance S from the train runs at this maximum velocity of `10 m//s` to catch the train at the same moment at which the train starts. Find the critical distance `S_(c)` for which passenger will take the ten seconds time to catch the trainA. 50 mB. 35 mC. 30 mD. 25 m
A particle is moving along X-axis under a force such that its position -time graph is as shown in figure. As the particle passes position (6) :A. it is moving along positive X-direction with a speed that is increasing with timeB. it is moving along positive X-direction with a speed that is decreasing with timeC. it is moving along negative X-direction with a speed that is increasing with timeD. it is moving along negative X-direction with a speed that is decreasing with time
A particle is moving along X-axis under a force such that its position -time graph is as shown in figure. As the particle passes position (5) :A. it is instantaneously at rest and will now move along positive X-directionB. it is moving along positive X-direction with a speed that is decreasing with timeC. it is moving along negative X-direction with a maximum speedD. it is moving along negative X-direction with a minimum speed
A particle is moving along X-axis under a force such that its position -time graph is as shown in figure. As the particle passes position (1):A. it is moving along negative X-direction with a speed that is increasing with timeB. it is moving along positive X-direction with a speed that is decreasing with timeC. it is moving along negative X-direction with a speed that is decreasing with timeD. it is moving along positive X-direction with a speed that is increasing with time
A particle is moving along X-axis under a force such that its position -time graph is as shown in figure. As the particle passes position (2):A. it is moving along negative X-direction with a maximum speedB. it is moving along positive X-direction with a minimum speedC. it is moving along negative X-direction and its speed is zero hereD. it is moving along negative X-direction with a minimum speed
A body is released from the top of a tower of height h. It takes t sec to reach the ground. Where will be the ball after time `(t)/(2)` sec?A. `(h)/(2)`B. `(h)/(4)`C. `(3h)/(4)`D. `(3h)/(2)`
In the question number 67, the time taken by the ball to reach the ground isA. 2 sB. 3 sC. 5 sD. 7 s

A point moves linearly with deceleration which is given by `dv//dt=-alphasqrt(v)`, where alpha is a positive constant. At the start `v=v_(0)`. The distance traveled by particle before it stops will be

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment