Question

A particle is moving in the xy plane in a sinusoidal course determined by y = A sin kx, where k and A are constant. Velocity of the particle in x-direction is constant and equal to v₀. Magnitude of the tangential acceleration at point x = is –

• A. Ak²v₀²
• B. Zero
• C. Ak²v₀
• D. Ak₀v₀²

Answer 1

Answer: (B)

Zero

Answer 2

x = v₀t

= x + y

$\overrightarrow { \mathrm { v } }$ = $\frac { \overrightarrow { \mathrm { dr } } } { \mathrm { dt } }$ = $\frac { \mathrm { dx } } { \mathrm { dt } }$ +

= + Akv₀ cos v₀ t k

= – Ak²v₀² sin v₀t k

at x = ,

= = v₀

= – Ak²v₀²