Equations of Motion by Calculus Method

First Equation of Motion:

Acceleration is defined as

    

                                                      ------ (1)

When time    0, velocity     (say); When time    t, velocity     (say)

Integrating equation (1) within the above limits of time and velocity, we get

      

        

      

     

                                    ------ (2)

Second Equation of Motion:

Velocity is defined as 

 

           

                                ------ (3)

when time    0, distance travelled    0; when time    t, distance travelled    s (say)

Integrating equation (1) within the above limits of time and distance, we get

      

      

        

      

                             ------ (4)

Third Equation of Motion:

By the definitions of acceleration and velocity,

  

      ×

      ×                               ------ (5)

When time    0, velocity    , distance travelled    0; When time    , velocity    , distance travelled     (say)

Integrating equation (5) within the above limits of velocity and distance, we get

   

   

      

            

         

                                                    ------ (6)

Fourth Equation of Motion:

By definition of velocity,

             

           

                                            ------ (7)

When time    (n − 1) second, distance travelled    (say); When time     second, distance travelled    (say)

Integrating equation (7) within the above limits of time and distance, we get

             

   

         

      

      

       

Problems:

1. A driver takes 0.20 s to apply the brakes after he sees need for it. If he is driving a car at speed of 54 km/h and the brakes cause deceleration of 6 ms-2, find the distance travelled by car after he sees need to put brakes.

Solution:

During the time of reaction, the car continues to move with uniform speed of 54km/h or 15 m/s.

∴ Distance covered during 0.20 s      15 × 0.20

          3.0 m

For motion with deceleration:

    15 m/s,     0,     − 0 m/s

As                                      

∴                                        

Or                                                        

        

       Total distance travelled    3.0 + 18.75

         21.75 m.

 

2. A body covers 12 m in  second and 20m in  second. How much distance will it cover in 4 seconds after the  second?

Solution:

Here                   12 m,     20 m

As                      

∴                     

or                                                                    ------ (1)

Also                   

or                                                                     ------ (2)

Subtracting (1) from (2),

    2a    8      or      a    4 ms-2

from (1),

    

         12 − 6 

            6 ms-1

Distance covered in 4 seconds after  second

                                           [Here ]

           

            216 − 80

            136 m.

 

3. Two buses A and B are at positions 50 m and 100 m from the origin at time t    0. They start moving the same direction simultaneously with uniform velocity of 10 m/s and 5 m/s respectively. At what position does A overtake B?

Solution:

Here we use equation of motion for constant velocity in cartesian form.

Given         (0)    50 m, (0)    100 m,

         10 m/s,    5 m/s

The position of the two buses at any instant t are 

(t)    (0) + t

  50 + 10t

(t)    (0) + t

  100 + 5t

When A overtakes B

(t)    (t)

     50 + 10t    100 + 5t

Or                        5t    50

        t    10 s

         (10)    (10)

              150 m.