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Asked by purushkumar91422 | 20th Apr, 2021, 09:00: 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 Thiyagarajan K | 20th Apr, 2021, 10:04: AM
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