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Particle Motion

A particle Q is moving +Y axis. Another particle P is moving in XY plane along a straight line x=-d (d>0) with a uniform speed v parallel to that of Q. At time t=0, particles P and Q happen to be along X axis whereas a third particle R situated at x=+d starts moving opposite to P with a constant acceleration a. At all further instants, the three particles happen to be collinear. Then Q

(A) has an initial speed v/2

(B)will come to rest after a time interval v/a

(C)has an acceleration –a/2

(D) will return to its initial position after a time interval 2v/a

Discussion:

$$y_2-y_1=\frac{y_2-y_1}{x_2-x_1}(x_2-x_1)$$
$$ y-vt=\frac{\frac{-1at^2}{2}-vt}{2d}(x+d)$$
On differentiating  taking x=0
$$v’-v=\frac{-at-v}{2d}(d)$$

At t=0,
$$ v’-v=\frac{0-v}{2}….(i)$$.
$$\Rightarrow v’=\frac{v}{2}$$
For $$v’=0,$$

we have,
$$ 0-v=\frac{-at-v}{2}$$
$$ \Rightarrow t=\frac{v}{a}$$
Differentiating (i)
$$ a’-0=\frac{-a}{2}$$
$$\Rightarrow a’=\frac{-a}{2}$$

Hence, the correct options will be a,b,c,d.

June 25, 2017

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