Find its acceleration vector?

Question

A particle moves in the xy-plane so that at any time t its coordinates are x=t^2 and y=4-t^3. Find its acceleration vector at t=1. 

 

Answer: (2, -6)

 

Please show all your work. 

Answer 1

x = t^2 ; y = 4-t^3

 

The first derivate, or velocity is; dx/dt = 2t ; dy/dt = -3t^2

 

The second derivate, or acceleration is; d²x/dt = 2 ; d²y/dt = -6t

evaluation for t = 1

 

x = 2(1) = 2     ;    y = -6(1) = -6

 

so:  (2, -6)

 

Charles G.

Answer 2

x(t)=t^2, y(t)=4-t^3

 

Velocity vector in the xy plane is the following:

 

v(t)= x'(t)i + y'(t)j

 

x'(t)= 2t       y'(t)= -3t^2

 

So v(t)= 2t i+ -3t^2 j

 

Acceleration vector in the xy plane is the following:

 

a(t)= x"(t)i + y"(t)j

 

x"(t)=2        y"(t)= -6t

 

So a(t)= 2 i + -6t j

 

At t=1, a(1) = 2 i + -6(1) j = 2 i + -6 j

 

 

Answer 3

I am assuming that you are in a calc based physics. You should know that the derivative of position is velocity and the derivative of velocity is acceleration. first derivative is (2t,-3t^2) the second derivative is (2,-6t) when you plug in 1 you get (2,-6).