I done the problem and was wondering if my answer is right I got 1.87 and then 12.4 degrees above x axis please help
At time t = 0 s, a puck is sliding on a horizontal table with a velocity 4.60 m/s, 32.0° above the +x axis. As the puck slides, a constant acceleration acts on it that has the following components: ax = –0.340 m/s2 and ay = –0.940 m/s2. What is the velocity of the puck at time t = 2.00 s?
To find the velocity of the puck at time t = 2.00 s, we need to use the equations of motion. The velocity of an object can be calculated using the formula:
v = u + at
Where:
v = the final velocity
u = the initial velocity
a = acceleration
t = time
In this problem, we are given:
u = 4.60 m/s (initial velocity)
ax = -0.340 m/s^2 (acceleration in the x-direction)
ay = -0.940 m/s^2 (acceleration in the y-direction)
t = 2.00 s (time)
To calculate the final velocity in the x-direction (vx), we need to consider the constant acceleration acting in the x-direction:
vx = ux + axt
Substituting the given values:
vx = 4.60 m/s + (-0.340 m/s^2) * 2.00 s
vx = 4.60 m/s - 0.68 m/s
vx = 3.92 m/s
To calculate the final velocity in the y-direction (vy), we also need to consider the constant acceleration acting in the y-direction:
vy = uy + ayt
Substituting the given values:
vy = 0 m/s + (-0.940 m/s^2) * 2.00 s
vy = 0 m/s - 1.88 m/s
vy = -1.88 m/s
Now, we can calculate the magnitude and direction of the final velocity using the Pythagorean theorem:
v = sqrt(vx^2 + vy^2)
v = sqrt(3.92 m/s)^2 + (-1.88 m/s)^2
v = sqrt(15.36 m^2/s^2 + 3.53 m^2/s^2)
v = sqrt(18.89 m^2/s^2)
v ≈ 4.35 m/s
To find the angle above the x-axis, we can use the inverse tangent function:
θ = atan(vy / vx)
θ = atan(-1.88 m/s / 3.92 m/s)
θ ≈ -26.5°
However, since the initial velocity was given as 32.0° above the +x axis, we need to add this angle to the angle we calculated:
θ = -26.5° + 32.0°
θ ≈ 5.5°
Therefore, the velocity of the puck at time t = 2.00 s is approximately 4.35 m/s, 5.5° above the +x axis.