Question

A particle moves along the x axis from the origin. The magnitude of the position vector at time t is

Answer 1

1) The average velocity is $-2.1\cdot 10^5 m/s$

2) The instantaneous velocity is $64t-260t^3$

Explanation:

1)

The average velocity of an object is given by

$v=\frac{d}{t}$

where

d is the displacement

t is the time elapsed

In this problem, the position of the particle is given by the function

$x(t) = 32t^2 - 65t^4$

where t is the time.

The position of the particle at time t = 6 sec is

$x(6) = 32(6)^2 - 65(6)^4=-83,088 m$

While the position at time t = 12 sec is

$x(12)=32(12)^2-65(12)^4=-1,343,232 m$

So, the displacement is

$d=x(12)-x(6)=-1,343,232-(-83,088)=-1,260,144 m$

And therefore the average velocity is

$v=\frac{-1,260,144 m}{12 s- 6 s}=-2.1\cdot 10^5 m/s$

2)

The instantaneous velocity of a particle is given by the derivative of the position vector.

The position vector is

$x(t) = 32t^2 - 65t^4$

By differentiating with respect to t, we find the velocity vector:

$v(t) = x'(t) = 2\cdot 32 t - 4\cdot 65 t^3 = 64t - 260 t^3$

Therefore, the instantaaneous velocity at any time t can be found by substituting the value of t in this expression.

Learn more about velocity:

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