# Doubt in kinematics (equations of motion)

Suppose a paticle travels from $$A$$ to $$D$$ inscribing a path like the one shown in the picture taking time $$t$$. My question is if I apply the equation of motion, which is $$s=ut + \frac 1 2at^2$$ where $$u$$ is the initial velocity and $$v$$ is the final velocity, will I get the length of path covered by the body(distance) or the length from $$A$$ to $$D$$(displacement) ? I know that in formulas, $$s$$ stands for displacement. But how can i get displacement which is $$A$$ to $$D$$ if I apply the time $$t$$ in the formulae which is the time taken by the body to inscribe the path $$ABCD$$ ?

(It is assumed the particle is traveling with uniform acceleration)

• Neither. The acceleration is not constant for the whole process and the SUVAT equations only work for constant accelerations both in magnitude and direction. However, you can break the problem in three different parts for which acceleration is constant and find out the final coordinates. – Mitchell May 28 '19 at 14:50
• I am sorry. I forgot to add constant acceleration in the question. I am editing the body – Ali May 28 '19 at 14:51
• You can't just assume constant acceleration. Particles moving with constant acceleration go in straight lines or in parabolas. You can't go around that horseshoe shape with uniform acceleration. – jacob1729 May 28 '19 at 15:03
• @Ali Not only as jacob 1729 said, but the acceleration at points B and C would appear to be infinite. – Bob D May 28 '19 at 15:27
• Hi and welcome to physics.SE! Please do not post formulae as plain text, but use MathJax instead. – ACuriousMind May 28 '19 at 16:14

will I get the length of path covered by the body(distance) or the length from A to D (displacement) ?

Your displacement "s" is the distance between A and D in a straight line, not following the path ABCD. This means that the formula you have, $$s=ut+\frac{at^2}{2}$$ (Remember to use MathJax to format equation whenever possible, so that it is easier to read) will obtain the net displacement, rather than the total distance traveled.

But how can I get displacement which is A to D if I apply the time 't' in the formulae which is the time taken by the body to inscribe the path ABCD ?

Remember that when calculating displacements you need the total time it took you to travel from A to D. From what you said, $$t$$ is precisely the total time, which is what the total displacement wants. Utilizing the total time and the displacement you can find the average velocities and accelerations, which is what I think you have in that formula.

Also, you are refering to a $$v$$ that does not appear in your question, try to keep an eye on that.

An object moving on a non-straight path has two accelerations. One is longitudinal (along the tangent to the path) and it changes the speed of the particle, and the other is transverse (perpendicular to the path) that changes the direction of motion.

When tracking the total distance traveled along the path if the speed is increasing linearly with time (constant acceleration), then yes the total distance traveled can be described by $$s(t) = s_0 + v_0 t + \frac{1}{2} a_{\rm along} t^2$$.

Note that the acceleration along the path $$a_{\rm along}$$ does not describe the total acceleration state of the object. In your case, the transverse acceleration around the sharp corners is going to be infinite.

You will get the length $$A$$ to $$D$$ because $$S$$ refers to displacement. Displacement is a vector, and when you add multiple vectors, in this case $$\vec{AB} + \vec{BC} + \vec{CD}$$, you will get $$\vec{AD}$$. Example $$\vec{AB} = x\, \mathrm{cm}$$. $$\vec{BC} = y \,\mathrm{cm}$$. $$\vec{CD} = x\,\mathrm{cm}$$.

$$\vec{AB} + \vec{BC} + \vec{CD} = x + y - x= y= \vec{AD}$$

(assuming $$\vec{BC} = \vec{AD}$$ where the path simulates a rectangle)