A small ball is moving on the flat surface (without rotation) such tha

1.	A small ball is moving on the flat surface (without rotation) such that its position is given by:where ex and ey are two unit vectors along the axis x and y. How can ı find the angle between velocity v(t) and the acceleration a(t) as t = 1?
Answer» Given that the position vector of the moing ball at t th instant is given by r(t)= t.e_x^-sin(t).ey Differentiating wrt t we get velocity vector v(t)= r'(t)= 1.e_x-cos(t).e_y Hence the angle of velocity vector with the positive direction of x axis at t=1 is in 4th quadrant and the angle of inclination is tan^-1cos(1) Now the accleration at t th instant is a(t)=v'(t)=sin(t).e_y Hence acceleration vector is directed towards +ve direction of y axis . So angle beween accleration and velocity vector at t=1 is given by π/2+tan^-1cos(1)

A small ball is moving on the flat surface (without rotation) such that its position is given by:where ex and ey are two unit vectors along the axis x and y. How can ı find the angle between velocity v(t) and the acceleration a(t) as t = 1?

Answer»

Given that the position vector of the moing ball at t th instant is given by

r(t)= t.e_x^-sin(t).ey

Differentiating wrt t we get velocity vector

v(t)= r'(t)= 1.e_x-cos(t).e_y

Hence the angle of velocity vector with the positive direction of x axis at t=1 is in 4th quadrant and the angle of inclination is

tan^-1cos(1)

Now the accleration at t th instant is

a(t)=v'(t)=sin(t).e_y

Hence acceleration vector is directed towards +ve direction of y axis .

So angle beween accleration and velocity vector at t=1 is given by

π/2+tan^-1cos(1)

Discussion

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