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Only the formulae given in (c), (d) and (f) are correct for describing the motion of the particle over the time-interval t1 to t2. The formulae given in (a), (b) and (e) are incorrect as from the slope of the graph we can see that the particle is not moving with constant acceleration.
As the speed is increasing in the time interval t = 0 s to t = 5 s the acceleration is positive and can be given by Speed at t = 2 s is Speed at t = 5 s is v1 = 15 m s-1  t1 = 5 - 2 = 3 s  Distance travelled in interval t = 2 s to t = 5 s is s1 Acceleration is negative after t = 5 s but has the same magnitude a2 = -2.4 m s-2 Speed at t =  5 s is u2 = 12 m s-1 t2 = 6 - 5 = 1 s Distance...
The distance traversed by the particle equals the area under the speed time graph The area under the curve is The particle has travelled a distance of 60 m from t=0 s to t=10 s. The average speed over this interval is
As both the stones are being accelerated due to gravity its effect will come on the relative motion only when one of them reaches the ground. Till that point of time, the relative velocity of the second stone would remain the same with respect to the first stone. Let us consider the upward direction to be positive. V1 = 15 m s-1 V2 = 30 m s-1 Vrel = V2 -V1 = 30 - 15 = 15 m s-1 Initial velocity...
The distance between the parents is s = 50 m The relative velocity of the child with respect to both his parents remains the same as the parents are also standing on the moving belt. v = 9 km h-1 = 2.5 m s-1 Time taken by the child in (a) and (b) is t = s/v = 20 s. As both, the parents are also standing on the belt as well the speed of the child would appear to be 9 km h-1 to both the parents...
(a) Speed of the child when the boy is running in the direction of the motion of the belt = 9 + 4 = 13 km h-1
Let us consider the upward direction to be positive Initial velocity of the ball (u) = 49 m s-1 The speed of the ball will be the same when it reaches the boy's hand's but will be moving in a downward direction. Therefore final velocity (v) = -49 m s-1 Acceleration (a) = -9.8 m s-2 Using the first equation of motion we have In the second case, as the ball has been thrown after the lift has...
Initial velocity u = 0 Acceleration, a = 1 ms-2 t = n seconds Let the total distance travelled in n seconds be Sn Similarly, total distance travelled in n - 1 second would be Sn-1 Distance travelled in nth second would be given as xn = Sn - Sn-1 As we can see the dependency of xn on n is linear we conclude the plot of the distance covered by the vehicle during the nth second versus n would...
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