#### Two trains 121m and 99m in length are running in opposite directions with velocities 40km/hr and 32km/hr. They will completely cross each other at

V_{A} = 40km/hr

V_{B} = -32km/hr

l_{A} = 121m

l_{B} = 99m Where l_{A} + l_{B} are lengths of train A and B.

Relative velocities of two trains will be
V_{AB }= V_{A} - V_{B }= 40 - (-32) = 72 km/hr = 20m/s.
Total distance travelled = 121 + 99 = 220 m.

Therefore, Time = 220/20 = 11 sec

#### Two parallel rail tracks run from North to South. On one track, train A is moving from South to North with a speed of 20 m/s. On the other track, train B is moving from North to South with a speed of 30 m/s. The relative velocity of A with respect to B is ____.

Let us consider the track to be along the Y-axis.

Then V_{A} = 20 m/s and V_{B} = -30 m/s.

Relative velocity of A with respect to B is
V_{AB} = V_{A} - V_{B}
= 20 - (-30) = 50 m/s (South to North)

#### The magnitude of resultant vector is equal to sum of the magnitudes of the two vectors and its direction is same as that of two given vectors.

This is as per the rule of geometric addition of vectors.

#### The motion of a train along a straight railway track is an example for

The straight line motion just specifies the distance travelled w.r.t time.

#### Among the following which of the two are vectors?

The two types of vectors are polar and axial vectors.

#### A particle moves with a constant speed v along a circular path of radius r and completes the circle in time T. The acceleration of the particle is ____.

For circular motion, v = 2πr/T ⇒r = vT/2π

Acceleration a=v^{2}/r=v^{2}/ (vT/2π) = 2πv/T

#### If the magnitude of sum of two vectors is equal to magnitude of difference of two vectors, the angle between these two vectors is

When the magnitude between the vectors sum and difference is equal; the vectors will lie perpendicular to each other. Hence θ = 90°

#### A man is at a distance of 6m behind a bus. The bus begins to move with a constant acceleration of 3 m/s^{2}. In order to catch the bus the minimum speed with which the man should run towards the bus is

Let us assume that the man will catch the bus after ’t’ seconds.

Then, the distance that the bus would have travelled will be

Distance travelled by the man is S_{2} = vt

To catch the bus, the passenger should cover the distance S_{1 }+ d = S_{2}
d+ 1/2 at^{2} = vt

⇒t = v± √( v^{2}-2ad)/a

't' can be real only if v ≥ √2ad

Therefore, the minimum speed that the man should run will be √2ad m/s = √ (2 × 3 × 6) = 6 m/s.

#### Two vectors of the same physical quantity are equal if

Property of equal vectors specifies its magnitude and direction that for two vectors to be equal, their magnitude and direction should be the same.

#### At 10 am, the position of an aircraft relative to an airport O is (220i + 50j) km, i and j being unit vectors due east and due north respectively. The velocity of the aircraft is (150i - 100j) km/hr. At 11 am, the aircraft will be at ____ from the airport.

It is given that at 10 am, the position of an aircraft relative to an airport O is (220i +50j)km. The velocity of the aircraft is (150i - 100j) km/hr.

Hence, in one hour, the displacement of the aircraft is (150i - 100j) km.

The new position is (220i + 50j) + (150i - 100j) = 370i - 50j km from the airport.

Distance of the aircraft from the airport = **√**(370^{2}+50^{2})≃373.36km.

#### Identify the vector quantity from the following physical quantities

Moment of spin of a body will have both magnitude and direction. It also obeys vector laws of addition. Hence that is the only vector quantity.

#### A particle is given a displacement of 4m in the x - y plane. If the x component of displacement vector is 2m, then y component is

D^{2} = Dx^{2} + Dy^{2}
(4)^{2} = (2)^{2} + Dy^{2}

Therefore, Dy^{2 }= 4^{2 }- 2^{2} = 16 - 4 = 12m

Dy = √12 = √4 ×√3 = 2 √3m.

#### If number of vectors is represented in magnitude and direction by the sides of a polygon taken in same order, then their resultant is represented in magnitude and direction by?

It is as per the polygon law of vectors addition which states that the resultant vector should be the closing side of polygon, but the direction will be opposite.

#### The magnitude of the resultant of two equal vectors is equal to the magnitude of either vector. Then the angle between the two vectors is

Because the magnitude of resultant vector is equal to the magnitude of given two vector (i.e.) R = A = B

If the two vectors are inclined at an angle 'θ' then, R^{2} = A^{2} + B^{2 }+ 2AB cosθ. (Because R=**√**A^{2}+B^{2}+2ABcosθ....) A^{2} = A^{2} + B^{2} + 2A^{2} cosθ)

A^{2} = 2A^{2} + 2A^{2} cosθ

1 = 2(1 + cosθ)

2 + 2 cosθ = 1

2 cosθ = –1

cosθ=−1/2=120°

#### Vectors can be added, subtracted and multiplied by

only laws of vectors is applicable to do vector operations.

#### n

#### A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. This implies that its ____?

A force as mentioned in the question makes the particle move along a circular path at a constant speed. Hence, kinetic energy is constant. But, velocity and acceleration continuously change due to change in direction.

#### Four bombs of 5 kg, 15 kg, 25 kg and 45 kg are fired from cannon with the same velocity in the same direction. Which of the following options is right ____? Note: Ignore air resistance

The time of flight does not depend upon the mass. Hence, all bombs will reach the ground in the same time.

#### A body is projected horizontally from a point above the ground. The motion of the body is given by the equations x = 2t and y = 5t^{2} where x and y are horizontal and vertical displacements in m at time t. The trajectory of the body is ____. 5 km/ hr

It is given that x = 2t and y = 5t^{2}.

Substituting for t in the equation for y, we get

y= (5×x/2)^{2}

=5x^{2}/4

i.e. y∝x^{2}

Therefore, the path is parabolic.

#### A car starting from rest at a constant acceleration covers a distance’s’ in time interval’t’. It covers a distance s_{2} in the next time interval at the same acceleration. The statement which is true is

Initial velocity is zero.
v = u + at. Therefore, v = at

∴ S_{1}= ut + 1/2 at^{2} = 12 at^{2}

For the next distance covered, u = at - v (from the previous motion)

∴ S_{2}= ut + 1/2at^{2} = 3/2 at^{2}

Hence, S_{2} = 3 S_{1}

#### A man stands at a distance’d’ m behind a bus which moves forward with a constant acceleration 'a'. He starts running at a constant speed 'v' m/s. The minimum speed of the passenger so that he can catch the bus is

Let us assume that the man will catch the bus after ’t’ seconds.

Then, the distance that the bus would have travelled will be S_{1}= 1/2 at^{2} ∵u = 0

Distance travelled by the man is S_{2} = vt

To catch the bus, the passenger should cover the distance S_{1} + d = S_{2}, d+ 1/2 at^{2 }= vt ⇒t = (v±** √**v^{2}-2ad)/(a) 't' can be real only if v ≥ √2ad

Therefore, the minimum speed that the man should run will be √(2ad) m/s

#### Scalar is specified by number and units. Here the number represents its

Because Scalar has only magnitude.

#### If the direction of a vector is along the axis of rotation, then the vector indicates anticlockwise or clockwise ____.

This is as per the definition of rotational effect.

#### Zero vector or null vector is a necessity because

for any vector calculation, vector cannot be nullified and hence there should be zero vector.

#### A boat moves from a point A (4î+5ĵ) to another point across the river. The new position of the boat is given by B (−7î−9ĵ). The displacement vector is given by ___.

Position A of the boat is given by 4î+5ĵ

Position B of the boat is given by −7î−9ĵ

We know that the displacement vector Δr→=Δxî+Δyĵ
⇒Δr→= (4î+5ĵ) − (−7î−9ĵ) ⇒Δr→=11î+14ĵ

#### If the quantity, position of a particle is represented using + and - signs then, the position is being expressed in

The signs indicate just the position of the place

#### A swimmer swims in a still water with a speed of 5m/s. While crossing the river his average speed is 3m/s. If he crosses the river in the shortest possible time, the speed of flow of water is

The resultant velocity of the swimmer V is perpendicular to velocity of water V_{W}.

Then V^{2} + V_{W}^{2} = V_{S}^{2}
V_{W}^{2} = VS^{2 }- V^{2} = 25 - 9 = 16.

Therefore V_{W} = Speed of water is 4m/s.

#### In the given diagram. R is the resultant of A and B. Then R = B/√2, value of angle θ is

As per the right angled triangle the angle will be 45°.

#### Vectors which are having equal or unequal magnitudes and are acting along parallel straight lines are,

As per the definition of collinear vectors.

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