You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Constrained Motion : Intuitive Approach

A train movin...

Question

A train moving with constant velocity 'V', a carriage on a road parallel to the track. at distance 'd', from it is observed to move so as to appear always in line with a more distant fixed object whose distance from the track is 'D'. What is velocity of the carriage?

V (1 - \frac{d}{D})

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

d \frac{V}{2} (1 - \frac{d}{D})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

V (1 - \frac{2 d}{D})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

V (1 - \frac{d}{2 D})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $V (1 - \frac{d}{D})$
$y = D t a n θ$
$x = (D - d) t a n θ$
$\frac{d x}{d t} = v_{1}$
$\frac{d y}{d t} = V$
From above equtions
$\frac{v_{1}}{V} = \frac{D - d}{d}$
$v_{1} = V (1 - \frac{d}{D})$

Suggest Corrections

Similar questions

The driver of a train moving with a constant velocity v₁along a straing track sights another train at a distance d ahrad of him on the same track moving in a same direction with a constant speed V2. He at once applies a brakes and gives his train a constant retardation f. There will be a collosioc of yhe trains if......?

a) v1 > V2 amd (V1 -V2)²/2f < d

b) V1<V2 and (V1+V2)² /2f > d

c) V1 > V2 and ( V1-V2)²/2f > d

d) V1>V2 and (V1²- V2²)/2f > d

If a car moves with velocity 'u' and its maximum retardation is 'a', its minimum stopping distance is proportional to

a)u

b)u²

c)u³

d)u^1/2

Two cars 1 and 2 move with velocities $v_{1} a n d v_{2}$ respectively on a straight road in the same direction. When the cars are separated by a distance d, the driver of the car 1 applies brakes and the car moves with uniform retardation $a_{1}$ . Simultaneously, the car 2 starts accelerating with $a_{2}$ . If $v_{1} > v_{2}$ , find the minimum initial separation between the cars to avoid collision between them.

Q. The driver of a train moving with a constant speed

v_{1}

along a straight track sights another train at a distance d ahead of him on the same track moving in the same direction with a constant speed

v_{2}

. He at once applies the brakes and gives his train a constant retardation f. There will be a collision of the trains if:

Q. An express train is moving with a velocity

v_{1}

. Its driver finds another train is moving on the same track in the same direction with velocity

v_{2} (v_{2} < v_{1})

. To escape collision, driver applies a retardation

a

on the train. The minimum time to escape the collision is

Join BYJU'S Learning Program

A train moving with constant velocity 'V', a carriage on a road parallel to the track. at distance 'd', from it is observed to move so as to appear always in line with a more distant fixed object whose distance from the track is 'D'. What is velocity of the carriage?

The correct option is A V(1−dD)y=Dtanθx=(D−d)tanθdxdt=v1dydt=V From above equtionsv1V=D−ddv1=V(1−dD)

The correct option is A $V (1 - \frac{d}{D})$
$y = D t a n θ$
$x = (D - d) t a n θ$
$\frac{d x}{d t} = v_{1}$
$\frac{d y}{d t} = V$
From above equtions
$\frac{v_{1}}{V} = \frac{D - d}{d}$
$v_{1} = V (1 - \frac{d}{D})$