SOLUTION

Speed is a scalar quantity and so, it cannot be negative.

SOLUTION

SOLUTION

When the velocity of a moving particle changes with time, the particle is said to have acceleration.
Thus, acceleration is the quantitative description of the rate of change of velocity with time.

SOLUTION

Distance travelled in last second = 0 + (2t -1)g/2  (using the formula of distance travelled in nth second)
Distance travelled in first three second = 0 + 1/2(g x 3²)   (using the 2nd equation of kinematics)
According to the question,

SOLUTION

u = - 20 m/s;a=9.8 m/s²; t= 5 second; S=?,
From the 2nd equation of kinemtaics: S= ut + (1/2 )at² 
  = - 20x5+ (1/2 ) x 9.8x (5)²  
  = 22.5 m

SOLUTION

Velocity, v = (dx/dt) = 12+8t
When t = 2 s,
v = 12+ (8x2) = 28 m/s

SOLUTION

SOLUTION

From the 1st equation of kinematics, 
v = u + at 
  = 20+4x6
  = 44 m/s

SOLUTION

 Relative velocity of motorcycle with respect to car=(80-65)=15 km/hr.

SOLUTION

SOLUTION

S = ut + ( 1/2 )at²
A body starts from rest so u=0
 s= ( 1/2 ) x 20 x 8² = 640 cm  

SOLUTION

The food packet has an initial velocity of 12 m/s in upward direction,  From first equation of motion, 
  v= - u + gt
or,= 12 +(9.8 x 2) = 7.6 m/s

SOLUTION

From the 3rd equation of kinematics, we get
 

SOLUTION

From the 3rd equation of kinematics,

SOLUTION

SOLUTION

Distance traveled=actual path traversed=  r Displacement =distance between the starting and end point =2r

SOLUTION

The distance traveled in first three seconds of its motion is, h = ( 1/2 )at²   h = ( 1/2 ) x 9.8 x 3² = 44.1 m
The distance traveled in last second of motion, S = 0 + ( 2t - 1 )( g/2 ) = ( 2t - 1 )4.9 It is given here that h=S  44.1 = ( 2t - 1 )4.9 So, t = 5 seconds

SOLUTION

Distance= area under v-t graph  is added up ,(without considering negative sign)
So, Distance = 4x2 + 2x2 + 2x2 = 16 m
Displacement=area under v-t graph is added up ,(considering the negative sign)
So, Displacement = 4x2 - 2x2 + 2x2 = 8m

SOLUTION

SOLUTION

Average acceleration = ( 0 + at )/2 = ( at )/2 
Putting this value in first equation of motion v = u + at
 v = u + [ ( at )/2 ] x t or, v = u + [ ( at² )/2 ]

SOLUTION

SOLUTION

SOLUTION

The speed of an object falling freely under gravity depends upon its height from which it is allowed to fall and not upon its mass. Since the paths are frictionless and all the objects are falling through the same vertical height, therefore their speeds on reaching the ground must be same. So, ratio of their speeds is 1:1:1  

SOLUTION

SOLUTION

Final velocity of bus, 
From the 1st equation of kinematics
v = u + at
So, 20 = 0 + ( a x 20 ) = 20a
Hence, a= 20/20 = 1 ms^-2

SOLUTION

Velocity of car = ( ds/dt ) = 12 + 6t - 6t²
So, the velocity of car at start (when t=0)=12+0+0=12m/s  

SOLUTION

SOLUTION

SOLUTION

SI unit of acceleration is m/s² cgs unit of acceleration is cm/s²

SOLUTION

For part AB   From 3rd equation of motion ,v² = u² – 2gs
 0 = u² – 2g(H/2) = u² – gH or, H = u²/g = 10²/ 10 =10 m

SOLUTION

 The only force acting on the ball is gravity. The ball will ascend until gravity reduces its velocity to zero and then it will descend. We will find the time the ball takes to reach its maximum height and then double the time to cover the round trip.       Using 1st euqation of kinematics 
   v= u + at = u – gt, we get:     or,  0 m/s = 50 m/s – (9.8 m/s²) t       Therefore, t = (50 m/s)/(9.8 m/s²) = 5.1 s       This is the time taken by the ball to reach its maximum height. The total round trip time is 2t = 10.2 s.

SOLUTION

The relative velocity of rain w.r.t. car is inclined to the vertical in the backward direction. 

Therefore, it will strike the front screen.

SOLUTION

For a falling body,
h =ut +  1/2 gt²

SOLUTION

  From A to B 
Using the 2nd equation of kinematics
      s = ut + 1/2 at²  2 = u x 2 + ( 1/2 ) x a x 2 x 2 or, 1 = u + a..........(1) Similarly from A to C 4.20 = u x 6 + (1/2) x a x 6 x 6 or,  0.7 = u + 3a.....(2) From (1) and (2), we get   2a = - 0.3 or a = - 0.15 ms^-2 u = 1 - a = ( 1 + 0.15 )ms^-1 = 1.15 ms^-1 Velocity at t = 9 sec, v = 1.15 - 0.15 x 9 = 1.15 -1.35 = - 0.2 ms^-1

SOLUTION

Let u and v be the speed of the two bodies respectively.   
Given  u + v = 6.....(1)
and        u - v = 4.....(2)
Adding (1) and (2), we get
   2u = 10 
or, u = 5 m/s
From eq(1), 
     5 +v = 6
or,      v =1 m/s
                       

SOLUTION

Relative velocity of approach = 60 - (-30) = 90 km/h.
Now at t= 0, the distance between the trains = 90 km
Therefore, the trains will collide in time = (90 km) / (90 km/h) =  1 hour

SOLUTION

Total distance covered = v₁t₁ + v₂t₂ = 1 x 60 + 3 x 60 = 240 m Total time taken = t₁ + t₁ = 60 + 60 = 120 seconds Thus, Average speed = 4 x 60 / 120 = 2 m/s. 

SOLUTION

Instantaneous velocity, v = dx/dt = 0 + 2 + 6t    Instantaneous acceleration, = a = dv/dt = 6 unit.

SOLUTION

By figure the velocity of parrot w.r.t. train is = 5–(–10) = 15m/sec So, time taken to cross the train is = ( length of train )/ (relative velocity) = 150 / 15 =10 s

SOLUTION

SOLUTION

SOLUTION

Distance travelled by a particle in half revolution is the half of the circumference i.e. (2r) / 2 = r

SOLUTION

Relative velocity of B w.r.t. A = v_BA = v_B - v_A = 5 - 4 = 1 km/h
Distance of B ahead A in time 't' = lv_ABl x t = 1 x 3 = 3 km

SOLUTION

SOLUTION

SOLUTION

By IInd equation of kinematics,              S = ut + ( 1/2 )at²     200 = u x 2 + ( 1/2 ) x a x 2²  u + a = 100 .....(1) The body covers 200 + 220 = 420 cm distance in 2 + 4 = 6 seconds.                   s = ut + ( 1/2 )at² or,              420 = u x 6 + ( 1/2 ) x a x 6² or  6( u + 3a ) = 420       u + 3a = 70 ....(2) On solving equation (1) and (2), a = –15 cm/s² and u = 115 cm/s. Velocity of the body at the end of 7th second v = u + at = 115 + (–15) × 7   = 115 – 105 = 10 cm/s

SOLUTION

Both the gun and the stone will reach earth at same time as both have initial velocity zero and dropped from same height.      t = ( 2h/g )^1/2
or, t (1/g)^1/2  

SOLUTION

Velocity decides the direction of the motion of a body. The acceleration simply tells us the rate of the change of velocity. For example, when a body is thrown vertically upward, its direction of velocity is upwards, that is why the body goes upward, where as its acceleration is downwards.

SOLUTION

Distance is a scalar quantity which has only magnitude and no direction, so it can be never negative.

SOLUTION

Speed is defined as the actual distance travelled by a body per unit time. Distance travelled can never be zero or negative but is always positive.
Hence, speed is either zero or positive, i.e.,
 
But speed can never be negative.

SOLUTION

Speed is defined as the actual distance travelled by a body per unit time. Distance travelled can never be zero or negative but is always positive. 
Velocity is defined as the displacement per unit time interval. Displacement is a vector quantity and can be positive, negative or zero.

SOLUTION

In the study of linear motion, we consider only the acceleration but not the rate of change of acceleration in a given motion of the object because basic laws of motion involve only acceleration and not the rate of change of acceleration. Example,
Force = mass x acceleration
         = mass x rate of change of velocity.

SOLUTION

 Let ‘a’ be the uniform retardation in both the cases.
 From the 3^rd equation of kinematics,  v² = u² + 2as        0 = u² +2(-a)s

     or, s µ u²

SOLUTION

SOLUTION

Here, S = 20.0 m.
From the 3^rd equation of kinematics,
v² = u²+2aS
For upward motion,
      0 = u²+2 x (-10) x 20
or, u = 20 m/s
If the total time of flight of the ball in going up and coming back, then total displacement in time t=0, i.e., S=0
From the 2^nd equation of kinematics,
     S = ut + ½ at²
or, 0 = 20 t + ½ (-10)t²       0= 20t -5t²
either t = 0 or,   t= 4.0 s
But time can never be zero; so, t = 4.0 s   Time interval of each ball = 4/4 = 1.0 s.

SOLUTION

As the speed is given by slope of x-t graph; hence,

SOLUTION

From the 2^nd equation of kinematics
       s = ut + ½ at²
Time taken by stone to fall, 

SOLUTION

Displacement is the shortest distance travelled by a body between the initial point and the final point. Here, net displacement is zero as Shipra returned back home.

SOLUTION

Let the body be dropped from the top of a tower of height ‘h’.
 Then, u = 0; acceleration a = g (acceleration due to gravity).
 Let y = vertical distance through which the particle falls in time t, the three equations of motion are written as,
 (a) v= u + at
 or, v = gt.... (1)
 (b) v² = u² +2as
 or, v² = 2gy...(2)
 (c) s = ut + ½ at²
 or, y = ½ gt²..(3)
 v = gt; v² = 2gy; y = ½ gt².

SOLUTION

Displacement is the shortest distance travelled by a body between the initial point (say A) and the final point (say B). For example 

SOLUTION

SOLUTION

Let u be the initial velocity of the body. Then  h = - ut₁ + (1/2)gt₁²    ... (1) and h = ut₂ + (1/2)gt₂²    ... (2) From eqs. (1) and (2), we get  - ut₁ + (1/2)gt₁² = ut₂ + (1/2)gt₂² or  (1/2)g x (t₁² - t₂²)= u ( t₁ + t₂ ) or (1/2)g x ( t₁ + t₂ )(t₁ - t₂) = u ( t₁ + t₂ )   or u = (1/2)g x (t₁ - t₂)   ...(3) Substituting the value of u in eq. (2) we get  h= (1/2)g x (t₁ - t₂)t₂ + (1/2)gt₂²  or  h= (1/2)g x t₁t₂  .........(4) For third body, h= (1/2)g x t² ...(5) From (4) and (5) (1/2)g x t² = (1/2)g x t₁t₂  t = ( t₁t₂ )^1/2

SOLUTION

Let car C be moving in opposite direction to cars A and B, with velocity  relative to the ground. Then the relative velocity of car C relative to A and B will be 
 

SOLUTION

The displacement-time graph is a straight line inclined to time axis upto time t_o, indicates a uniform velocity. After time t₀ the displacement-time graph is a straight line parallel to time axis indicates the particle is at rest.

SOLUTION

Moon’s gravity is 1/6th of the gravity of Earth. 
So, the balloon will fall with acceleration = (g/6) ms^-2.

SOLUTION

Graph (d) is not possible, because at a particular time t, displacement cannot have two values.

SOLUTION

Istantaneous velocity = dx/dt = d(2 + 4t + 5 t² )/dt = 4 + 10t

SOLUTION

The speedometer of a vehicle records instantaneous speed because it does not indicate the direction of motion. Velocity is a vector quantity and is associated with magnititude as well as direction.

SOLUTION

SOLUTION

Since the person as well as the bag fall down the same vertical height and experiencing same acceleration due to gravity; hence, they acquire same speed for reaching the ground.

SOLUTION

Given, s = 3t³ + 7t² + 5t +8
  velocity, v = ds/dt = 9t² + 14t +5
  acceleration, a = dv/dt = 18t +14
  acceleration, a_(t=1) = 18 + 14 = 32 m/s².

SOLUTION

u = 0,
v = 180 km/h 
   

SOLUTION

A particle thrown up has zero velocity at its uppermost point but its acceleration is acceleration due to gravity i.e 9.8 ms^-2, acting vertically downwards.

SOLUTION

Velocity-time graph will be parallel to time axis.

SOLUTION

No. Distance is the total length covered between two points by a body which can never be zero.
Whereas if the body returns back to the same initial point, then dispalcement is zero.

SOLUTION

The magnitude of the average velocity of a particle is equal to the average speed if it moves with constant velocity in a straight line.

SOLUTION

Yes. In Uniform circular motion, body has constant speed and acceleration is towards centre

SOLUTION

Both will stop at the same distance.

SOLUTION

The Co-ordinate axes with respect to which all measurements are done is called a frame of reference.

SOLUTION

Yes; if direction changes continuously.

SOLUTION

Scalar quantities are the quantities which have magnitude only but vector quantities are those which have magnitude and direction. Vectors follow triangle law E.g. mass, time ,speed are scalar and force, acceleration are vector quantities.

SOLUTION

Speed is the ratio of distance and time Whereas velocity is the ratio of displacement with respect to time.
Speed is a scalar quantity whereas Velocity is a vector quantity.

SOLUTION

1. Displacement is the shortest distance travelled by a body between two points. Its unit is same as that of distance i.e. metre.
2. The displacement of body can be positive, negative or zero.
3. Displacement of a body is vector quantity.
4. Displacement of a body can be equal or less than the distance covered between same points.

SOLUTION

The total distance covered by the body will be twice of the distance of path AB=2x, where as the displacement covered by body will be zero. As AB =-BA so the total displacement will be zero.

SOLUTION

1. A body travels equal displacement in equal intervals of time.
2. Velocity in uniform motion does not depend on the choice of origin.                              
3. Velocity in uniform motion does not depend on time interval.                                 
4. For uniform motion in straight line in same direction the magnitude of displacement is equal to magnitude of distance.

SOLUTION

A body is said to have uniform velocity, if it covers equal displacements in equal intervals of time. A body is said to be moving with variable velocity if either speed or direction or both changes with time. Velocity at particular instant of time is said to be instantaneous velocity, the time interval tends to zero.

SOLUTION

SOLUTION

Let distance AB= X km. 

Given, speed = 50 km/h
Time taken from A to B = t₁

      t₁ = (X/50) hr

Time taken from B to A
         t₂ = (X/40) hr

Average speed = Total distance/Total time

                    =  (X + X)/[ (X/50) + (X/40)]

                    = ( 2X / 9X ) x 200

                         = ( 400/9 ) km/hr

Average velocity  = Total displacement / Total time

                              =  0.

SOLUTION

When object moves, its overall magnitude of velocity vector remains same. A simple translation does not change the magnitude of the velocity vector but changes the values of its components. For example, during projectile motion magnitude of resultant velocity remains the same. Similarly, in the case of rotation about a fixed origin, the components of the vector will get modified, because the angles changes but, the length of the vector remains the same.

SOLUTION

The acceleration vector is independent of time; thus, the particle is moving with constant acceleration in the x-y plane. But velocity changes as the x-components of the velocity vector is time-dependent.

SOLUTION

\begin{array}{l}\text{For a man standing on the pavement the direction of relative velocity remains}\\ \text{same}\text{. However, relative velocity of the rain with respect to running man changes}\text{.}\\ \text{let us assume the velocity of the running man is}\overrightarrow{v_{\text{m}}}\text{and the velocity of the rain}\\ \text{is}\overrightarrow{v_{\text{r}}}\text{then relative velocity of the rain with respect to the running man is,}\\ \overrightarrow{{\text{v}}_{\text{rm}}}\text{=}\overrightarrow{{\text{v}}_{\text{r}}}\text{+}\left(\overrightarrow{{\text{-v}}_{\text{m}}}\right)\\ \overrightarrow{{\text{v}}_{\text{rm}}}\text{=30}{\text{ms}}^{\text{-1}}\text{+ (-10}{\text{ms}}^{\text{-1}}\text{) = 20}{\text{ms}}^{\text{-1}}\\ \text{The direction of relative velocity of the rain with respect to the running man is}\\ \text{tan}\text{θ}=\frac{10}{30}=0.33\\ \Rightarrow \text{θ}=18°\end{array}

SOLUTION

SOLUTION

SOLUTION

SOLUTION

SOLUTION

The balls A and B will reach the ground simultaneously. It is because vertical motions of the balls are identical and the balls are coming down under the influence of acceleration due to gravity only. However, the ball A strikes the ground just at the foot of the tower while the ball B will hit the ground quite away from the foot of the tower.

SOLUTION

SOLUTION

SOLUTION

SOLUTION