Question

6.
A particle moving with uniform acceleration in a straight
line has a velocity v ms1 at a position x metre from
origin is given by v = ✓196 - 16x. Its displacement
after 2 sec is​

Answer 1

relation between position and velocity is given as v = √{196 - 16x}

we know, velocity is the rate of change of position with respect to time.

i.e., v = dx/dt

or, dx/dt = √{196 - 16x}

or, ∫dx/√{196 - 16x} = ∫dt

let 196 - 16x = P²

differentiating both sides,

-16 dx = 2 P dP => dx = -P dP/8

so, ∫dx/√{196 - 16x} = ∫PdP/-8P =(-1/8) ∫dP

= -1/8 P

= -1/8 √(196 - 16x)

so, $\left[\frac{-1}{8}\sqrt{196-16x}\right]^x_0=[t]^2_0$

or, -1/8 [√(196 - 16x) - 14]= 2

or, -√(196 - 16x) = 16

or, √(196 - 16x) -14 = -16

or, 196 -16x = -2

or, x = 198/16 = 12.375 m

hence, position of particle at t = 2sec is 12.375 m