A particle is travelling on the x-axis. Its velocity time graph is given:
Which of the following graphs represent the same motion?
A particle is travelling on the x-axis. Its velocity time graph is given:
Which of the following graphs represent the same motion?
The correct options are
A
B
C
D
Lets break the motion and analyze.
Part - I
1. ∵ Velocity time is a straight line uniformly increasing so its acceleration is constant.
Also if we see v = mt ← equation of line. Y = mx + c
So we see acceleration is constant
Slope at each point is increasing so is the velocity.
2. Corresponding displacement time graph will be increasing.Option (a)
3.till here distance will be same as displacement.
Interval t1−t2
1.Speed time graph will be same
2.Since velocity is constant throughout
i.e., acceleration of the particle is 0.
3.Displacement will increase linearly as slope is constant throughout and so is the velocity.
4. Distance time graph will be same.
Interval t2−t3
1.Speed time graph will be same.
2.Velocity is decreasing linearly so body is acted upon by constant deceleration.Also relation of v to t is like be v = -t + c.dvdt−k = acceleration has to be negative and constant so
3. clearly it can be seen that slope of displacement time graph is decreasing and becoming 0 at t3 so is velocity.
4. Distance time graph will be same till here.
Interval t3−t4
1.Speed cannot bve negative as speed is a scalar so whatever velocity the particle has the speed will be same but positive.
2.just like previous case velocity is decreasing at a constant rate so same acceleration time graph.
3.
Displacement initially slope = 0. Velocity = 0, later the tangent keeps becoming steeper i.e., slope is increasing but since angle is obtuse so the velocity is increasing in negative direction.So its correct.
4. as with time displacement decrease the distance should increase simultaneously.
Interval t4−t5
Velocity
1.Speed cannot be negative.Speed will have same valve but positive.
2.velocity in increasing linearly so acceleration is constant.
3.
The tangent is steeper in the starting and becomes sleeping in the end so the slope is decreasing.But since θ is obtuse so velocity is negative.So the velocity is becoming less negative with time which is exactly the velocity time curve.
4.
Distance can't be negative it will keep on increasing than decreasing.
A
B
C
D
Lets break the motion and analyze.
Part - I
1. ∵ Velocity time is a straight line uniformly increasing so its acceleration is constant.
Also if we see v = mt ← equation of line. Y = mx + c
So we see acceleration is constant
Slope at each point is increasing so is the velocity.
2. Corresponding displacement time graph will be increasing.Option (a)
3.till here distance will be same as displacement.
Interval t1−t2
1.Speed time graph will be same
2.Since velocity is constant throughout
i.e., acceleration of the particle is 0.
3.Displacement will increase linearly as slope is constant throughout and so is the velocity.
4. Distance time graph will be same.
Interval t2−t3
1.Speed time graph will be same.
2.Velocity is decreasing linearly so body is acted upon by constant deceleration.Also relation of v to t is like be v = -t + c.dvdt−k = acceleration has to be negative and constant so
3. clearly it can be seen that slope of displacement time graph is decreasing and becoming 0 at t3 so is velocity.
4. Distance time graph will be same till here.
Interval t3−t4
1.Speed cannot bve negative as speed is a scalar so whatever velocity the particle has the speed will be same but positive.
2.just like previous case velocity is decreasing at a constant rate so same acceleration time graph.
3.
Displacement initially slope = 0. Velocity = 0, later the tangent keeps becoming steeper i.e., slope is increasing but since angle is obtuse so the velocity is increasing in negative direction.So its correct.
4. as with time displacement decrease the distance should increase simultaneously.
Interval t4−t5
Velocity
1.Speed cannot be negative.Speed will have same valve but positive.
2.velocity in increasing linearly so acceleration is constant.
3.
The tangent is steeper in the starting and becomes sleeping in the end so the slope is decreasing.But since θ is obtuse so velocity is negative.So the velocity is becoming less negative with time which is exactly the velocity time curve.
4.
Distance can't be negative it will keep on increasing than decreasing.