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# Solve this question

Asked by purushkumar91422 20th April 2021, 9:00 AM
In first one second period, toy-car's velocity is changed from 0 to 6 m/s .

Hence acceleration a is determined by equation of motion " v = u + a t "
where v is velocity after t second and initial velocity is u

Hence acceleration a = v/t = 6 / 1 = 6 m/s2

Distance S1 travelled in first one second is calculated using equation of motion

" S = ut + (1/2) a t2 "

S1 = (1/2) × 6 × 1 × 1 = 3 m

After one second field is reversed , hence acceleration direction is reversed but its magnitude is same

Hence acceleration = -6 m/s2

Since the toy car is moving with velocity 6 m/s at end of 1 second from initial time,
time taken to reduce its velocity to zero is determined as follows

Equation of motion is   " v = u - a t  "

In above equation of motion v = 0 , hence  t = u / a = 6 /6 = 1 s

Distance moved in time duration from t =1 s to t = 2 s is calculated as follows .

S2 = u t - (1/2) a t2 =  ( 6 × 1 ) - [  (1/2) × 6 × 1 × 1 ] = 3 m

Since it  is given that toy car has moved for 2 seconds after electric field reversed.

We have  seen that it has moved in forward direction 1 second after field reversal.

Hence for one second period it will move in backward due to change in direction of acceleration.

Distance S3 moved in backward direction for 1 second time duration is calculated as follows

S3 = - (1/2) × 6 × 1 × 1 = -3 m

Average velocity = Net displacement / time = ( S1 + S2 + S3 ) / t  = ( 3 + 3 - 3 ) / 3 = 1 m/s

Average speed = Total distance / time = ( S1 + S2 + | S3 | ) / t  = ( 3+3+3 ) / 3 = 3 m/s
Answered by Expert 20th April 2021, 10:04 AM
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