Question

The displacement of a particle is described by $x=t+2t^2+3t^3-4t^4$. The acceleration at t=3 sec is

• A. +378 units
• B. +374 units
• C. -378 units
• D. -374 units

Answer 1

The correct option is D -374 units

Displacement $x=t+2t^2+3t^3-4t^4$

Velocity: The rate of charge of displacement

Velocity$=\frac{dx}{dt}=t+4t+9t^2-16t^3$

The rate of charge of velocity in acceleration

$\therefore$ Acceleration$=a=\frac{dv}{dt}=\frac{d}{dt}\left[\frac{dx}{dt}\right]=\frac{d^{2}x}{dt}$

$\therefore \frac{d^{2}x}{d{t}^2}=4+18t-48t^2$

Acceleration $t=3$ is

$\frac{d^{2}x}{d{t}^2}$ at $t=3$

$=4+18t-48t^2$

$=4+18\left(3\right)-48(3{)}^2$

$=-86$ units.