Question

A puck is moving on an air hockey table. Relative to an x, y coordinate system at time t = 0 s, the x components of the puck's initial velocity and acceleration are v0x = +2.9 m/s and ax = +5.7 m/s2. The y components of the puck's initial velocity and acceleration are v0y = +9.3 m/s and ay = -5.2 m/s2. Find (a) the magnitude v and (b) the direction θ of the puck's velocity at a time of t = 0.50 s. Specify the direction relative to the +x axis.

Answer 1

The kinematics allows finding the answer for the velocity and in angles for a time t = 0.5 s they are:

         a)  $v_y$ = 6.7 m /s

        b)  θ = 49.4º

Given parameters

• Initial velocity        v₀ₓ = 2.9 m /s and $v_{oy}$ = 9.3 m / s

• The acceleration   aₓ = 5.7 m /s² and $a_y$ = -5.2 m / s²

To find

     For t = 0.5 s

• a) Speed vetical

• b) The angle

Kinematics studies the motion of bodies, giving relationships between position, velocity and acceleration.

The reference system is a coordinate system with respect to which to carry out all the measurements, it is very important in all the problems, in this exercise the reference system is given  the veloicity and the accelerations are given.

a) Let's find the velocity for t = 0.5 s

We use the kinematics relation

            v = v₀ + a t

x- axis

            vₓ = v₀ₓ + aₓ t

            vₓ = 2.9 + 5.7 0.5

            vₓ = 5.75 m / s

y-axis

            $v_y = v_o_y + a_y \ t$

            $v_y$ = 9.3 -5.2 0.5

            $v_y$ = 6.7 m / s

b) To find the direction of the velocity, we use trigonometry

          tan θ = $\frac{v_y}{v_x}$

          θ = tan⁻¹ $\frac{v_y}{v_x}$

          θ = tan⁻¹ $\frac{6.7}{5.75}$

          θ = 49.4º

 

In conclusion using the kinematics relations we can find the answer for the velocity and in angles for t = 0.5 s they are:

        a) $v_y$vy = 6.7 m /s

        b) θ = 49.4º

Learn more about kinematics here:

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