Here is a question in kinematics which is a popular one and therefore requiring your attention:
The two ends of a train running with a constant acceleration passes a certain point with velocities v1 and v2. The velocity with which the middle point of the train passes the point is
(a) (v1+v2)/2 (b) √(v1 2 +v22) (c) (v12+v22)/2 (d) (v1+v2)/√2
The two ends of a train running with a constant acceleration passes a certain point with velocities v1 and v2. The velocity with which the middle point of the train passes the point is
(a) (v1+v2)/2 (b) √(v1 2 +v22) (c) (v12+v22)/2 (d) (v1+v2)/√2
(e) √[(v12+v22)/2]
If ‘s’ is the length of the train, the velocity of the train changes from v1 to v2 when it moves through the distance ‘s’. Therefore we have,
v22- v12 = 2as from which a = (v22 - v12 )/2s
If’ ’v’ is the velocity with which the mid point of the train passes the reference point, we have v2 = v12 +2a(s/2). Substituting for the acceleration ‘a’ from the above equation, v = √[(v12 +v2 2 )/2].
The following question appeared in the Kerala Engineering Entrance Test paper of 2002:
A body dropped from a height ‘h’ with an initial velocity zero reaches the ground with a velocity 3km/hour. Another body of the same mass is dropped from the same height ‘h’ with an initial velocity 4km/hour. It will reach the ground with a velocity
(a) 3km/hour (b) 4km/hour (c) 5km/hour (d) 12km/hour (e) 8km/hour
We have, v2 = u2 + 2as with usual notations.
Therefore, 32 = 0 + 2gh for the first case and
v2 = 42 + 2gh for the second case. These two equations yield the value v= 5km/hour. Note that we did not convert the velocities into m/s since the answer is required in km/hour. [Note that the answer is independent of the masses of the bodies].
Now consider the following simple question:
A particle starting from rest travels with uniform acceleration for t1 seconds and then travels (continuously) with uniform retardation and comes to rest in another t2 seconds. If the total distance traveled is ‘s’, the maximum velocity attained during the motion is
(a) s/(t1 – t2) (b) s/(t1 + t2) (c) 2s/(t1 – t2) (d) 2s/(t1 + t2) (e) s/t1 + s/t2
Since the acceleration and retardation are uniform, this can be easily solved using the concept of average velocity. If ‘v’ is the maximum velocity attained, the average velocity during the acceleration part as well as the deceleration part is v/2.
Therefore, s = (v/2)t1+ (v/2)t2 = (v/2)(t1+t2) from which v = 2s/(t1+t2).
If ‘s’ is the length of the train, the velocity of the train changes from v1 to v2 when it moves through the distance ‘s’. Therefore we have,
v22- v12 = 2as from which a = (v22 - v12 )/2s
If’ ’v’ is the velocity with which the mid point of the train passes the reference point, we have v2 = v12 +2a(s/2). Substituting for the acceleration ‘a’ from the above equation, v = √[(v12 +v2 2 )/2].
The following question appeared in the Kerala Engineering Entrance Test paper of 2002:
A body dropped from a height ‘h’ with an initial velocity zero reaches the ground with a velocity 3km/hour. Another body of the same mass is dropped from the same height ‘h’ with an initial velocity 4km/hour. It will reach the ground with a velocity
(a) 3km/hour (b) 4km/hour (c) 5km/hour (d) 12km/hour (e) 8km/hour
We have, v2 = u2 + 2as with usual notations.
Therefore, 32 = 0 + 2gh for the first case and
v2 = 42 + 2gh for the second case. These two equations yield the value v= 5km/hour. Note that we did not convert the velocities into m/s since the answer is required in km/hour. [Note that the answer is independent of the masses of the bodies].
Now consider the following simple question:
A particle starting from rest travels with uniform acceleration for t1 seconds and then travels (continuously) with uniform retardation and comes to rest in another t2 seconds. If the total distance traveled is ‘s’, the maximum velocity attained during the motion is
(a) s/(t1 – t2) (b) s/(t1 + t2) (c) 2s/(t1 – t2) (d) 2s/(t1 + t2) (e) s/t1 + s/t2
Since the acceleration and retardation are uniform, this can be easily solved using the concept of average velocity. If ‘v’ is the maximum velocity attained, the average velocity during the acceleration part as well as the deceleration part is v/2.
Therefore, s = (v/2)t1+ (v/2)t2 = (v/2)(t1+t2) from which v = 2s/(t1+t2).
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