Question

The x and y coordinates of a particle at any time t are x = 5t - 3t2 and y = 5t respectively, where x and y are in meter and t in second. The speed of the particle at t = 1 second is​

Answer 1

Answer:

$v=\sqrt{26}~m/s$

Explanation:

Parametric Equation of the Velocity

Given the position of the particle at any time t as

$r(t) = (x(t),y(t))$

The instantaneous velocity is the first derivative of the position:

$v(t)=(v_x(t),v_y(t))=(x'(t),y'(t))$

The speed can be calculated as the magnitude of the velocity:

$v=\sqrt{v_x^2+v_y^2}$

We are given the coordinates of the position of a particle as:

$x=5t-3t^2$

$y=5t$

The coordinates of the velocity are:

$v_x(t)=(5t-3t^2)'=5-6t$

$v_y(t)=(5t)'=5$

Evaluating at t=1 s:

$v_x(1)=5-6(1)=-1$

$v_y(1)=5$

The velocity is:

$v=\sqrt{(-1)^2+5^2}$

$v=\sqrt{1+25}$

$\mathbf{v=\sqrt{26}~m/s}$