Kinematics Problem involving variable acceleration

AI Thread Summary
The discussion revolves around a kinematics problem involving a particle with variable acceleration. The particle starts with a constant velocity of 8 m/s in the x-direction and has a varying acceleration given by a = (7√t j) m/s². The participant initially struggled with integrating the acceleration to find the velocity but later clarified that the x-component of velocity remains constant at 8i. The correct expression for the velocity vector as a function of time is confirmed to be 8i + 4.67t^(3/2)j. The conversation highlights the importance of correctly identifying constant and variable components in kinematics problems.
zewei1988
Messages
21
Reaction score
0

Homework Statement


What if the acceleration is not constant? A particle starts from the origin with velocity 8 i m/s at t = 0 and moves in the xy plane with a varying acceleration given by vector a = (7√t j) m/s2, where t is in s. Answer the following using t as t as necessary.

Determine the vector velocity of the particle as a function of time.

Homework Equations





The Attempt at a Solution



I could find the x-component of the velocity vector, but integrating 8 with respect to time. However, when my answer is wrong when i integrated 7(t^0.5) in an attempt to find the velocity of the particle as a function of time.

4.67t^(1.5) is my answer.
 
Physics news on Phys.org
Welcome to PF!

Hi zewei1988! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)
zewei1988 said:
I could find the x-component of the velocity vector, but integrating 8 with respect to time.

What do you mean? The 8i velocity component is constant.
 
I'm sorry, I din integrate it with time. I mixed up my solutions for the other part of the question. I simply put 8 as the velocity vector along the x-axis.
 
zewei1988 said:
I'm sorry, I din integrate it with time. I mixed up my solutions for the other part of the question. I simply put 8 as the velocity vector along the x-axis.

So your answer is 8i + 4.67t3/2j ?

That looks ok to me. :confused:
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top