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: a = - 2 0.360 m/s and da, = -0.980 m/s. What is the velocity of the puck at time t = 1.50 s? 2 0​

Answer 1

The velocity of the puck at time t = 1.50 s is -0.272 m/s in the y-direction.

Step-by-step explanation:

To find the velocity of the puck at time t = 1.50 s, we need to use the equations of motion.

The horizontal component of the acceleration will not affect the puck's velocity in the y-direction, so we can ignore it.

Using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can calculate the y-component of the velocity:

vy = uy + ayt

Where:

• uy is the initial y-component of velocity (3.60 m/s * sin(35°))
• ay is the y-component of acceleration (-0.980 m/s²)
• t is the time (1.50 s)

Substituting the given values into the equation and solving for vy, we get:

vy = 3.60 m/s * sin(35°) + (-0.980 m/s²) * (1.50 s)

vy = -0.272 m/s

Therefore, the velocity of the puck at time t = 1.50 s is -0.272 m/s in the y-direction.