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The figure shows a velocity-time graph of a particle moving along a straight line
Choose the incorrect statement. The particle comes to rest at
- a)t = 0 s
- b)t = 5 s
- c)t = 8 s
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The figure shows a velocity-time graph of a particle moving along a st...
The particle comes to rest (v = 0) at two instants. It comes to rest first time at somewhere between 4th second and 6th second. At t = 8 second it again comes to rest.
We will find when the particle comes to rest for the first time. Between 4th and 6th second (time interval of 2 seconds) the particle changes it's velocity from 10 m/s to -20 m/s. So, the acceleration on the particle during this part of motion is,
a = (-20 - 10)/2 = -15 seconds
The particle comes to rest when v = 0
v - u = at
0 - 10 = (-15)t
⇒ t = 1.5 seconds
So the particle come to rest at (4 + 1.5)th = 5.5 th second and 8th second
We will find when the particle comes to rest for the first time. Between 4th and 6th second (time interval of 2 seconds) the particle changes it's velocity from 10 m/s to -20 m/s. So, the acceleration on the particle during this part of motion is,
a = (-20 - 10)/2 = -15 seconds
The particle comes to rest when v = 0
v - u = at
0 - 10 = (-15)t
⇒ t = 1.5 seconds
So the particle come to rest at (4 + 1.5)th = 5.5 th second and 8th second
Most Upvoted Answer
The figure shows a velocity-time graph of a particle moving along a st...
The particle comes to rest (v = 0) at two instants. It comes to rest first time at somewhere between 4th second and 6th second. At t = 8 second it again comes to rest.
We will find when the particle comes to rest for the first time. Between 4th and 6th second (time interval of 2 seconds) the particle changes it's velocity from 10 m/s to -20 m/s. So, the acceleration on the particle during this part of motion is,
a = (-20 - 10)/2 = -15 seconds
The particle comes to rest when v = 0
v - u = at
0 - 10 = (-15)t
=> t = 1.5 seconds
So the particle come to rest at (4 + 1.5)th = 5.5th second and 8th second
We will find when the particle comes to rest for the first time. Between 4th and 6th second (time interval of 2 seconds) the particle changes it's velocity from 10 m/s to -20 m/s. So, the acceleration on the particle during this part of motion is,
a = (-20 - 10)/2 = -15 seconds
The particle comes to rest when v = 0
v - u = at
0 - 10 = (-15)t
=> t = 1.5 seconds
So the particle come to rest at (4 + 1.5)th = 5.5th second and 8th second
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Community Answer
The figure shows a velocity-time graph of a particle moving along a st...
The particle comes to rest (v = 0) at two instants. It comes to rest first time at somewhere between 4th second and 6th second. At t = 8 second it again comes to rest.
We will find when the particle comes to rest for the first time. Between 4th and 6th second (time interval of 2 seconds) the particle changes it's velocity from 10 m/s to -20 m/s. So, the acceleration on the particle during this part of motion is,
a = (-20 - 10)/2 = -15 seconds
The particle comes to rest when v = 0
v - u = at
0 - 10 = (-15)t
⇒ t = 1.5 seconds
So the particle come to rest at (4 + 1.5)th = 5.5 th second and 8th second
We will find when the particle comes to rest for the first time. Between 4th and 6th second (time interval of 2 seconds) the particle changes it's velocity from 10 m/s to -20 m/s. So, the acceleration on the particle during this part of motion is,
a = (-20 - 10)/2 = -15 seconds
The particle comes to rest when v = 0
v - u = at
0 - 10 = (-15)t
⇒ t = 1.5 seconds
So the particle come to rest at (4 + 1.5)th = 5.5 th second and 8th second
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