Question

At time t = 0 s, a puck is sliding on a horizontal table with a velocity 3.60 m/s,
35.0° above the +x axis. As the puck slides, a constant acceleration acts on it
that has the following components: ax = –0.360 m/s2 and ay = –0.980 m/s2.
What is the velocity of the puck at time t = 1.50 s?

Answer 1

Well you're trying to find the v at 1.50. Which we'll call v2 brah. How does one get to v2 from the given though? Well you find the components of v1 first. v1 is 3.60 m/s as it is the first velocity we're dealing with. Hopefully you know how to find the components. You'll get 3.6sin35 and 3.6cos3.5 for v1y and v1x respectively. v1y is the y component of v1 and v1x is the x component. You'll get v1y as 2.06 and v1x as 2.95. Now use the kinematic equation v=vo+at. Only we're solving for v2's components now from v1's components. The equation becomes v2x=axt+v1x and v2y=ayt+v1y. Now we can use those accelerations they gave us. Set up your equations as follows. v2x=-.36(1.5)+2.95, and v2y=-.98(1.5)+2.06. You should get 2.41 and .59. Now just square both and root them. sqrt(2.41^2+.59^2), and you'll get v2. As doing this with any components of a vector will give you the vector's magnitude itself brah. You should get 2.48 as the result. To find the direction create a diagram for the components of v2 like you did v1. Instead of using 3.6 as the hypotenuse use 2.48, and set up your equation as cos(A)=v2x/v2, or sin(A)=v2y/v2, to get the angle. Plug in your prior numbers, cos(A)=2.41/2.48, and you'll get 13.6 when you do a cos inverse function on your calc.