0

Guest

Courses New

A particle of mass m moves on the x-axis as follows: it starts from rest at t = 0 from the point x = 0, and comes to rest at t = 1 at the point x = 1. No other information is available about its motion at intermediate times If α denotes the magnitude of instantaneous acceleration of the particle, then which of the following is/are true? (a) |α|cannot remain positive throughout (b)| α| cannot exceed 2 at any point in its path (c) |α|must be equal to 4 at some point or points in its path ​ (d) |α |must change sign during the motion, but no other assertion can be made with the information given Obviously the a part is correct because the the particle has to deaccelerate to come to a stop. i think one can do this by shm also but please give me the solution with the help of kinematics also. I dont understand why the c part is correct.

A particle of mass m moves on the x-axis as follows: it starts from rest at t = 0 from the point x = 0, and comes to rest at t = 1 at the point x = 1. No other information is available about its motion at intermediate times 
If α denotes the magnitude of instantaneous acceleration of the particle, then which of the following is/are true?
(a) |α|cannot remain positive throughout
(b)| α| cannot exceed 2 at any point in its path
(c) |α|must be equal to 4 at some point or points in its path
​(d) |α |must change sign during the motion, but no other assertion can be made with the information given

Obviously the a part is correct because the the particle has to deaccelerate to come to a stop.
i think one can do this by shm also but please give me the solution with the help of kinematics also. I dont understand why the c part is correct.

Grade:12

3 Answers

Narendra Pratap
19 Points
9 years ago
We know that the initial velocity of the particle is zero and the final  too. Considering constant magnitude of acceleration we can draw velocity-time graph for the motion. I am not able to draw it here, t will be an isosceles triangle with its base along time axis. Hieght of this triangle will be maximum velocity v and it will occur at 0.5sec.
for (a) part the magnitude will be same but direction will be opposite
(b) we can find acceleration by using v= u+at for increasing velocity part 
 v = 0 +a*0.5 implies that a = v/0.5, now v can be calculated by computing area under v-t graph. 
we know total distance coverd is 1m the area under the graph = 1
 
as graph is a isosceles triangle thus ½ v*1 =1  ( height of triangle will be v and base is 1sec) 
 
v = 2m/s  thus a 2/0.5  = 4m/s^2, 
 
thus option (b) is wrong but (c ) is correct.
Aslo for option (d) acceleration will change its direction in between the motion. 
 
Narendra Pratap
19 Points
9 years ago
We know that the initial velocity of the particle is zero and the final  too. Considering constant magnitude of acceleration we can draw velocity-time graph for the motion. I am not able to draw it here, t will be an isosceles triangle with its base along time axis. Hieght of this triangle will be maximum velocity v and it will occur at 0.5sec.
for (a) part the magnitude will be same but direction will be opposite
(b) we can find acceleration by using v= u+at for increasing velocity part 
 v = 0 +a*0.5 implies that a = v/0.5, now v can be calculated by computing area under v-t graph. 
we know total distance coverd is 1m the area under the graph = 1
 
as graph is a isosceles triangle thus ½ v*1 =1  ( height of triangle will be v and base is 1sec) 
 
v = 2m/s  thus a 2/0.5  = 4m/s^2, 
 
thus option (b) is wrong but (c ) is correct.
Aslo for option (d) acceleration will change its direction in between the motion. 
 
Narendra Pratap
19 Points
9 years ago
We know that the initial velocity of the particle is zero and the final  too. Considering constant magnitude of acceleration we can draw velocity-time graph for the motion. I am not able to draw it here, t will be an isosceles triangle with its base along time axis. Hieght of this triangle will be maximum velocity v and it will occur at 0.5sec.
for (a) part the magnitude will be same but direction will be opposite
(b) we can find acceleration by using v= u+at for increasing velocity part 
 v = 0 +a*0.5 implies that a = v/0.5, now v can be calculated by computing area under v-t graph. 
we know total distance coverd is 1m the area under the graph = 1
 
as graph is a isosceles triangle thus ½ v*1 =1  ( height of triangle will be v and base is 1sec) 
 
v = 2m/s  thus a 2/0.5  = 4m/s^2, 
 
thus option (b) is wrong but (c ) is correct.
Aslo for option (d) acceleration will change its direction in between the motion. 
 

Think You Can Provide A Better Answer ?

Other Related Questions on General Physics

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More