Question

A particle moves along a straight line as per equation x² = αt² + 2βt + γ, where x is the distance travelled and α, β, γ are constants. Its acceleration varies as

• A. x^-3
• B. x^3/2
• C. x^-2/3
• D. x²

Answer 1

Answer: (A)

x^-3

Answer 2

Given x² = (αt² + 2βt + γ)

∴ x = (αt² + 2βt + γ)^1/2

v = $\frac { \mathrm { dx } } { \mathrm { dt } } = \frac { 1 } { 2 }$ (αt² + 2βt + γ)^1/2 (2αt + 2β)

= (αt² + 2βt + γ)^1/2 (αt + β)

and a = $\frac { \mathrm { dv } } { \mathrm { dt } } = - \frac { 1 } { 2 }$ (αt² + 2βt + γ)^1/2

(2αt + 2β) (αt + β) + (αt² + 2βt + γ)^-1/2 (α)

= -(αt² + 2βt + γ)^-3/2 (αt + β)² + α(αt + 2β + γ)^-1/2

= (αt² + 2βt + γ)^-3/2 – (α²t² –  α² – 2αβt + α²t² + 2αβt + αγ)

= $\frac { \alpha \gamma - \beta ^ { 2 } } { \left( \alpha t ^ { 2 } + 2 \beta t + \gamma \right) ^ { 3 / 2 } } = \frac { \alpha \gamma - \beta ^ { 2 } } { x ^ { 3 } }$

Thus a ∝x^-3