The position vector of a particle changes with time according to the relation \(\rm \vec{r}(t)=15 t^{2} \hat{i}+\left(4-20 t^{2}\right) \hat{j} . \) What is the magnitude of the acceleration at t = 1?

CONCEPT:

Position Vector and Acceleration

• The position vector of a particle provides information about its location in space as a function of time.
• Acceleration is the rate of change of velocity with respect to time.
• To find acceleration, we need to differentiate the position vector twice with respect to time.

EXPLANATION:

• Given position vector: r(t) = 15t² i + (4 - 20t²) j
• First, we find the velocity by differentiating the position vector with respect to time:
  • v(t) = dr(t)/dt = 30t i - 40t j
• Next, we find the acceleration by differentiating the velocity with respect to time:
  • a(t) = dv(t)/dt = 30 i - 40 j
• At t = 1, the acceleration vector is:
  • a(1) = 30 i - 40 j
• To find the magnitude of the acceleration:
  • |a(1)| = √(30² + (-40)²)
  • |a(1)| = √(900 + 1600)
  • |a(1)| = √2500
  • |a(1)| = 50

Therefore, the correct answer is option 4: 50.