Solve Kinematics Problem: Find Distance Traveled Till t=5

AI Thread Summary
The discussion revolves around finding the distance traveled by a particle described by the displacement equation s = 6 + 12t - 2t^2 until t=5. Participants express confusion about interpreting the displacement equation, particularly regarding the meaning of s at t=0 and how to calculate total distance from t=0 to t=5. It is clarified that the equation provides the position of the particle over time, and the derivative ds/dt represents its velocity. To find the total distance, one must consider the particle's position at various time intervals, rather than just the displacement at t=5. The conversation highlights the frustration of grappling with complex kinematics problems despite thorough study.
babita
Messages
61
Reaction score
0

Homework Statement



Displacement of a particle moving in a straight line is
s= 6 + 12t - 2t^2
find the distance traveled till t=5.


Homework Equations



firstly i am having confusion in correctly interpreting the equation. The equation gives us the displacement in 't' th second.(correct?)
if i put t=0 in the eq i get s=0...now what does that mean? the particle is at x=6 at t=0?

at t=6.5 s=0 and before that s always comes out to be positive
as the particle is moving in straight line we can just add up the displacements till t=5 to get distance covered till t=5


The Attempt at a Solution



but how to add up displacments. the equation gives the displacement in t th sec.
like if i put t=5 in the equation i get s=16 but this is the displacement in 5th second only..i need the displacements in all time b/w 0 to 5 seconds
.
.
i also referred to the solution provided in the book which goes as follows:
ds/dt = v= 12-4t
( now i don't agree ds/dt would be the velocity...please explain)
 
Physics news on Phys.org
I suspect that by "displacement" they mean "distance from the origin". Then the equation

s= 6 + 12t - 2t^2

would in fact be the position of the particle with respect to time, and ds/dt would indeed be its velocity.
 
assuming that the answer comes correct...thanks...but i want to kill the author...it discourages if one can't solve a prob even after studying deeply :((
 
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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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