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Grade 12General Physics

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.

Profile image of Jai Mahajan
11 Years agoGrade 12
Answers icon

3 Answers

Profile image of Narendra Pratap
ApprovedApproved Tutor Answer11 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. 
 
Profile image of Narendra Pratap
11 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. 
 
Profile image of Narendra Pratap
11 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. 
 

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