4.8 Motion in a plane with constant acceleration

Suppose that an object is moving in x-y plane and its acceleration a is constant. Over an interval of time, the average acceleration will equal this constant value. Now, let the velocity of the object be v0 at time t = 0 and v at time t.

Then, by definition

Or,

(4.33a)

In terms of components :

(4.33b)

Let us now find how the position r changes with time. We follow the method used in the one-dimensional case. Let ro and r be the position vectors of the particle at time 0 and t and let the velocities at these instants be vo and v. Then, over this time interval t, the average velocity is (vo + v)/2. The displacement is the average velocity multiplied by the time interval :

(4.34a)

Or,

It can be easily verified that the derivative of Eq. (4.34a), i.e. NEETprep Audio Note (English):    

gives Eq.(4.33a) and it also satisfies the condition that at t=0, r = ro. Equation (4.34a) can be written in component form as

(4.34b)

One immediate interpretation of Eq.(4.34b) is that the motions in x- and y-directions can be treated independently of each other. That is, motion in a plane (two-dimensions) can be treated as two separate simultaneous one-dimensional motions with constant acceleration along two perpendicular directions. This is an important result and is useful in analysing motion of objects in two dimensions. A similar result holds for three dimensions. The choice of perpendicular directions is convenient in many physical situations, as we shall see in section 4.10 for projectile motion.

Answer From Eq. (4.34a) for r0 = 0, the position of the particle is given by

Therefore,

Given x (t) = 84 m, t = ?

5.0 t + 1.5 t2 = 84 ⇒ t = 6 s

At t = 6 s, y = 1.0 (6)2 = 36.0 m NEETprep Audio Note (English):    

At t = 6 s,

v = 23.0 î +12.0 ĵ 

speed

.