Question

A point moves according to the law x = at, y = at(1 - αt) where a and α are positive constants and t is time. Find the moment at which angle between velocity vector and acceleration vector is π/4.

• A. 1/α
• B. α
• C. 1/α²
• D. None of these

Answer 1

Answer: (A)

1/α

Answer 2

$\overrightarrow{v} = \frac{dx}{dt}\widehat{i} + \frac{dy}{dt}\widehat{j} = a\widehat{i} + (a - 2a\alpha t)\widehat{j}$

f = $\frac{dv}{dt} = - 2a\alpha\widehat{j}$

cos $\frac{\pi}{4} = \frac{- 2a\alpha\widehat{j}.\left\lbrack a\widehat{i} + (a - 2a\alpha t)\widehat{j} \right\rbrack}{2a\alpha\sqrt{a^{2} + (a - 2a\alpha t)^{2}}}$

⇒ $\frac{1}{\sqrt{2}} = \frac{- \lbrack a - 2a\alpha t\rbrack}{\sqrt{a^{2} + (a - 2a\alpha t)^{2}}}$ or

2(1 - 2αt)² = 1 + (1 - 2αt)²

or (1 - 2αt)² = 1 or t = 1/α