Question

A motor car takes 10 seconds to cover 30 meters and 12 seconds to cover 42 meters. Find theuniform acceleration of the car and its velocity at the end of 15 seconds.[10]​

Answer 1

Answer:

0.25 m/s^2      ;      3.75 m/s

Explanation:

 In the first 10 second:

    Distance traveled = 30 m

    Time taken = 10 s

  Thus, velocity is ( 30 / 10 ) m/s or 3 m/s.

In the first 12 second:

   Distance traveled = 42 m

   Time taken = 12 s

  Thus, velocity is ( 42 / 12 ) m/s

Time gap between the change of velocities is 12 s - 10 s = 2 s.

Therefore,

       Acceleration is ( chance in velocity ) / ( change in time )

⇒ { (42/12) - 3 } / 2  m/s^2

⇒ { ( 42 - 36 ) / 12 } / 2  m/s^2

⇒ ( 6 / 12 ) / 2  m/s^2

⇒ 6 / 24  m/s^2

⇒ 0.25 m/s^2 = acceleration( uniform ).

      Now, using v = u + at.

⇒ v( velocity till 15 s ) = 0 + ( 0.25 )( 15 )  m/s

⇒ v = 3.75 m/s

Answer 2

Answer:

Case I

Distance covered by car =30 metres

Time taken = 10 second

So as velocity =distance /time

velocity =30/10 = 3m/sec

Case II

Distance travelled =42 metres

Time taken =12 seconds

Then, velocity =distance /time

velocity =42 /12 =7/2=3.5 m/sec

Also, change in time =(12-10)seconds =2 seconds

Now as, Acceleration =[change in velocity /change in time]

Acceleration =(3.5-3)/2 ms¯²

Acceleration =0.5/2 ms¯²

Acceleration =0.25 ms¯².

Now, by using the first equation of motion,

v=u+at

Velocity in 1 second (till end) is displacement /1

In the same v=at =0.25×15 ms¯¹.

v = 0 +( 0.25)×(15)

v = 3.75 m/s