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Grade: 12
        

A particle starts from rest with a fconstant acceleration. At a time t second, the speed is found to be 100m/s and one second later the speed becomes 150m/s. find (a) the acceleration and (b) the distance travelled during the (t+1)th second.

9 years ago

Answers : (2)

Sachin Tyagi
31 Points
							

(a) Velocity at time t is 100m/s =a.(t second).............................1

and velocity at time (t+1) second is  150m/s=a. (t+1) second....................2

(b) consider the interval t second to (t+1) second, time elapsed =1s

Initial Velocity = 100 m/s

Final velocity = 150 m/s.

Thus,  (150 m/s)2 = (100m/s)2 + 2(50 m/s2 )x

or,                        x= 125 m.

9 years ago
bala krishna
33 Points
							This is a simply question that can be solved in more than one method.

Given,

Let velocity at the instant time 't' = 100 m/s
now velocity after 1 second that is at 't+1' sec velocity =150

Let acceleration be a. 

 now from equations of motion,
        v = u +at
      150 = 100 + at' ...............1

    where t' = [t+1]-t =1 sec.
      
there fore a = 50 m/s2.      [a bit] 
 
    distance travelled :

       v2 -u2 = 2as
       
there fore,2500 m
						
9 years ago
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