Solve Kinematics Problem: Find Distance Traveled Till t=5

In summary, the conversation discussed the equation for displacement of a particle moving in a straight line and how to find the distance traveled till t=5. There was some confusion about correctly interpreting the equation and understanding what it represents. The solution provided in the book was also discussed, with some disagreement about whether ds/dt represents velocity.
  • #1
babita
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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)
 
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  • #2
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.
 
  • #3
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 :((
 

1. How do I solve a kinematics problem to find the distance traveled until t=5?

To solve this type of problem, you will need to use the equation d = v0 * t + 1/2 * a * t^2, where d is the distance traveled, v0 is the initial velocity, a is the acceleration, and t is the time. You will also need to know the values for at least two of these variables.

2. What is the difference between distance and displacement in kinematics?

Distance is the total length of the path traveled, while displacement is the shortest distance between the starting and ending point. In other words, displacement takes into account the direction of motion, while distance does not.

3. Can I use kinematics equations to solve problems with non-constant acceleration?

Yes, kinematics equations can be used to solve problems with non-constant acceleration by using calculus or by breaking the motion into smaller intervals with constant acceleration.

4. How do I know which kinematics equation to use for a given problem?

There are four main kinematics equations that can be used to solve problems: d = v0 * t + 1/2 * a * t^2, v = v0 + a * t, v^2 = v0^2 + 2 * a * d, and d = (v0 + v)/2 * t. To determine which equation to use, you will need to identify the known variables and the variable you are trying to solve for.

5. What are some common mistakes to avoid when solving kinematics problems?

Some common mistakes to avoid when solving kinematics problems include using the wrong equation, not paying attention to the direction of motion, and not properly converting units. It is also important to double-check your calculations and make sure your answer makes sense in the context of the problem.

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