Kinematics

1  A train is moving with uniform velocity of 36kmh^-1 for 10s. Find the Distance travel by it.

     

                     Solution

Given Data

Velocity = v = 36kmh^-1  

             v = 36 × 1000/3600

             v = 10ms^-1

Time = t = 10 s

To find

 Distance = s = ?

Formula

             velocity = distance/time

             v = S/t

             S = v × t

Calculation

                   

As

        S = v × t

By putting the values

       

       S = 10 × 10 = 100 m

       S = 100m

Hence total distance will be 100m

2. A train starts from rest. It movies through 1 km in 100 s with uniform acceleration. What will be its speed at the end of 100 s.

Given Data

  Initial velocity = vi = 0 ms^-1

  Distance           = S = 1km = 1000m

  Time                  = t =100 s

To find

 Final Velocity   = vf =?

Formula

                                        S = v_it+1/2 at²

                                        v_f = v_i + at

             

Calculation

First of all we have to find “a” after getting the     value of “a” we have to find “v_f”.

                                                      S = v_it+1/2 at²

By putting the values we get,

                                                      1000 = (0) × 100 + 1/2 × (a) × (100)²

Opening the square of 100.

                                                     1000 =  1/2 × (a) × (10000)

By dividing 2.

                                                     1000 =   (a) × (5000)

By Dividing 5000 both sides

                                                          

                                                         1000/5000 =   a

                                                                                 a = 0.2 ms^-2

Now we have to calculate the value of “Final Velocity” like as,

                                   v_f = v_i + at

Putting the values we, get

                                  v_f = 0 + (0.2)×100

Multiplying 0.2 by 100

                             v_f = 20 ms^-1

Hence the body having the final velocity of

v_f = 20 ms^-1            

3. A car has velocity of 10ms^-1 for half minute with 0.2 ms^-2. Find the distance travel during this time and the final velocity?

Given Data

Initial velocity = vi =10ms^-1

Time = half Min. = 30 s

Acceleration = a = 0.2ms^-2

To find

Final velocity = vf =?

Distance   = S = ?

Formula

                             v_f = v_i + at

                      

Calculation

as

                      v_f = v_i + at

By putting the values

                               

                              v_f = 10 + (0.2)×30

                              v_f =16ms^-1

Hence the final  velocity of object will be 16ms^-1.

4. A tennis ball is hit vertically upward with a velocity of 30 ms^-1. It takes 3s to reach the highest point. Calculate the maximum height reached by the ball. How long it will take to return to ground?

Given Data

                        Initial velocity = v_i = 30 ms^-1

                        Time      = t = 3 s

                        Final velocity = v_f = 0 ms^-1

                        Acceleration = g = -10 ms^-2

To find

                        Total time of motion = t’ =?

                        Max. height = S =?

Formula

                         S = v_it+1/2 at²

Calculation

5. A car moves with uniform velocity of 40ms^-1 for 5 s. it comes to rest in the next 10 s with5. A car moves with uniform velocity of 40ms^-1 for 5 s. it comes to rest in the next 10 s with uniform deceleration. Find declaration? Total distance?

Motion with uniform velocity

Given Data

                        Velocity= vi  =  vf = v_av  = 40ms^-1

                                                Time  = t = 5s

To find

Distance S1 = ?

Formula

                        S = v_av×t

Calculation

                        S = 40×5 = 200m

Motion with variable velocity

Given Data

                        Initial velocity = vi =  40ms^-1

                        Final velocity = vf =  0ms^-1

                                                Time                = t = 10 s

To find

                        Deceleration = a = ?

                        Distance = S2 = ?

Formula

Calculation

6. A train starts form rest with an acceleration of 0.5 ms^-2 find its speed in kmh^-1, when it has moved through 100m.

Given Data

Initial velocity = v_i = 0 ms^-1

Covered Distance = S = 100 m

Acceleration = a  = 0.5 ms^-2

To find

                            Vf = ? kmh^-1

Formula

                           2as = v_f² – v_i² 

Calculation 

as

                          2as = v_f² – v_i² 

 Putting the values

                            2as = v_f² – v_i²

                            2 × 0.5 × 100 = v_f² – 0²

100 = v_f²

By taking square root to both sides we get,

10ms^-1 = v_f

For   kmh^-1

Multiplied by 3600/1000v_f = 10 ×3600/1000  

v_f = 36  kmh^-1 

Hence final velocity will be 36 kmh^-1

7. A train starting from rest accelerates uniformly and attains a velocity 48 kmh^-1 in two minutes. It travels at this speed for 5 minutes. Finally it movies with uniform retardation and I stopped after 3 minutes. Find the total distance traveled by the train.

                           

Solution

Part (a): A train starting from rest accelerates uniformly and attains a velocity 48 kmh^-1 in two minutes

Given Data

vi = 0ms^-1

vf = 48kmh^-1 = 48×1000/3600 = 40/3 ms^-1

t =2min = 120s

To Find

S_a=?

Formula

Calculation

Part (b) It travels at this speed for 5 minutes.

Given Data                      

v_f = v_i = v_av = 40/3

t = 5min = 5×60 = 300s

To find

                          S = ?

Formula

                         S = v_av × t

Calculation                  

                

                   S = v_av × t

                  Putting the values

                  S = 40/3 × 300 =4000m

Part (c) Finally it movies with uniform retardation and It stopped after 3 minutes.

 Find the total distance traveled by the train.

Given Data

             v_i = 40/3 ms^-1

             v_i = 0 ms^-1

             t = 3min = 3×60 = 180s

To find     

                 S_c = ?

Formula

Calculation

       

Total Distance

Total distance  = S_a + S_b + S_c

Total distance = 4000m+800m+1200m

Total distance = 6000m.

8.A cricket ball is hit vertically upward and returns to ground 6s later.

 Calculate (i) maximum height reached by the ball (ii) initial velocity of the ball.

Given Data

                

                  v_f = 0ms^-1

                  g = +10ms^-2 (for upwards motion)

                  g = -10ms^-2 (for downwards motion)

                  t₁ = 3s (for upwards motion)

                  t₂ = 3s (for downwards motion)

To find

                  h = S =?

                 v_i = ?

Formula

              

                   S = h = v_it+1/2 at²

                        v_f = v_i + at

Calculation

            As we know that

                                          v_f = v_i + at

Putting the values we get

                                          0 = v_i  + ( -10 )(3)

                                                 v_i  = 30ms^-

Similarly

             S = h = v_it+1/2 at²

_­Putting the values

When body moves downwards then its

Initial velocity will be zero (v_i = 0ms^-1)

     S = h = (0)(3)+1/2 (10)(3)²

           h = 45 m

Hence the initial velocity will be 30m/s and

maximum height will be 45m.

9.When brakes are applied, the speed of train decreases from 96 kmh^-1 to 48kmh^-1 in 800m. How much distance before coming to rest? (Assuming the retardation to be constant)

Given Data

v_i = 96kmh^-1 = 96 × 1000/3600  = 26.67ms^-1

v_f = 48kmh^-1 = 48 × 1000/3600  = 13.33ms^-1

S = 800 m

To find

                S = ?

Formula

               

            2aS = v_f² – v_i²

Calculation

as

            2aS = v_f² – v_i²

By putting the values

            2a(800) = (13.33)² – (36.67)²

                        a = -533.60/1600

                        a = -0.33ms^-1

when body comes to rest then

v­_I = 13.33ms^-1

v­_f = 0 ms^-1

as

                                 2aS = v_f² – v_i²

putting the values

                           2(-0.33)S = 0² – (13.33)²

                               -0.66S  = – (13.33)²

S = (13.33)²/0.66

S = 266m

Hence required distance will be 266m!

10. In the above problem find the time taken by the train

to stop after the application of breaks.

Given Data

            v­_f = 0 ms^-1

            v­_f = 26.67 ms^-1

            a = -0.33 ms^-2

To find

                        t = ?

Formula

                        v_f = v_i + at

Calculation

                         As

                          v_f = v_i + at

Putting the values

                           26.67 = 0 + (0.33 )t

                        t = 26.67/0.33

                           t = 80 s nearly

Hence the required time will be 80s.