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Grade: 10
        A stopwatch was started when a car traveling in a straight line crosses a fixed point. The car travels a distance of 200m in the first 2 seconds and 250m in the next 4 seconds. What will be the velocity of the car at the end of the fifth second?
11 months ago

Answers : (1)

Arun
15912 Points
							
 

let initial velocity = u & accleration is a then

 

distance travelled in 4 sec = S6

 

using , s=ut + at2/2

 

          S6 = 4u + a36/2

 

               =  6u+18a

 

distance covered in 2 sec = S2 = 2u + 4a/2

 

                                           =2u+2a = 200           (given)

 

                                       u+a = 100             ........................1

 

now distance covered in 2 to 6 sec is given 220cm

 

      S6-200 = 250

 

     6u + 18a = 450       

 

        u+3a = 75             .................2

 

from 1& 2

 

a = -25/2 & u=112.5

 

velocity after 5 sec = V

 

       V5 = u + at

 

           =112.5 - (25/2)*5=50cm/sec

 

 

Regards

Arun (askIITians forum expert)

11 months ago
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