A hockey puck with mass 0.160 kg is at rest at the origin (X = 0)a player applies a force of 0.250 N to the puck, parallel to the x- axis; he continues to apply this force until t = 2.00 s. a) What are the position and speed of the puck at t = 2.00 s? b)If the same force is again applied at t = 5.00 s, what are the position and speed of the puck at t = 7.00 s?

a = Fap/m = 0.25 / 0.160 = 1.56 m/s^2.

a. V = a*t = 1.56m/s^2 * 2s=3.125 m/s.
d = 0.5at^2 = 0.5*1.56*2^2 = 3.12 m.

To solve this problem, we will use Newton's second law of motion and the equations of motion to find the position and speed of the puck at different times.

a) To find the position and speed of the puck at t = 2.00 s, we first need to calculate the acceleration of the puck using Newton's second law.

Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the net force is 0.250 N and the mass of the puck is 0.160 kg.

F = m*a
0.250 N = 0.160 kg * a

Solving for acceleration:
a = 0.250 N / 0.160 kg
a = 1.563 m/s^2

Now, we need to find the initial velocity (u) of the puck, which is 0 m/s since it is at rest.

Using the equation: v = u + a*t, where v is the final velocity and t is the time interval.

v = 0 m/s + (1.563 m/s^2) * (2.00 s)
v = 3.126 m/s

To find the position (x) of the puck, we will use the equation: x = u*t + 0.5*a*t^2

x = (0 m/s) * (2.00 s) + 0.5 * (1.563 m/s^2) * (2.00 s)^2
x = 0 m + 0.5 * (1.563 m/s^2) * 4.00 s^2
x = 3.13 meters

Therefore, at t = 2.00 s, the position of the puck is 3.13 meters and its speed is 3.13 m/s.

b) To find the position and speed of the puck at t = 7.00 s, we need to calculate the time interval between t = 5.00 s and t = 7.00 s.

Time interval = t(final) - t(initial)
Time interval = 7.00 s - 5.00 s
Time interval = 2.00 s

Since the force applied is the same as before, and there are no other forces acting on the puck, the acceleration remains the same at 1.563 m/s^2.

Now, let's find the final velocity (v) using the equation v = u + a*t.

v = 3.126 m/s + (1.563 m/s^2) * (2.00 s)
v = 3.126 m/s + 3.126 m/s
v = 6.252 m/s

To find the position (x), we will use the equation: x = u*t + 0.5*a*t^2.

Note that the initial velocity (u) is the final velocity at t = 2.00 s.

x = 3.13 m/s * 2.00 s + 0.5 * (1.563 m/s^2) * (2.00 s)^2
x = 6.26 m + 0.5 * (1.563 m/s^2) * 4.00 s^2
x = 6.26 m + 3.126 m
x = 9.386 meters

Therefore, at t = 7.00 seconds, the position of the puck is 9.386 meters, and its speed is 6.252 m/s.

To determine the position and speed of the puck at different times, we need to use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by the acceleration of the object.

Let's break down the problem step by step:

a) We know that the player applies a force of 0.250 N to the puck, parallel to the x-axis for a duration of 2.00 s. Since the puck starts at rest at the origin (X = 0), we can determine its position and speed at t = 2.00 s.

Step 1: Calculate the acceleration of the puck.
Using Newton's second law, we can calculate the acceleration (a) of the puck:
a = F / m
where F is the force applied to the puck and m is the mass of the puck.
Plugging in the values:
a = 0.250 N / 0.160 kg

Step 2: Calculate the velocity of the puck.
To calculate the velocity (v) of the puck at t = 2.00 s, we can use the following equation:
v = u + at
where u is the initial velocity of the puck, which is 0 m/s since it starts from rest.
Plugging in the values:
v = 0 m/s + a * 2.00 s

Step 3: Calculate the position of the puck.
To calculate the position (X) of the puck at t = 2.00 s, we can use the following equation:
X = ut + 0.5at^2
where u is the initial velocity of the puck, t is the time, and a is the acceleration.
Plugging in the values:
X = 0 m/s * 2.00 s + 0.5 * a * (2.00 s)^2

b) Now, let's determine the position and speed of the puck at t = 7.00 s, after the same force is applied again at t = 5.00 s.

Step 1: Calculate the acceleration of the puck.
We'll use the same formula as before to calculate the acceleration of the puck.
a = F / m (force divided by mass)
Plugging in the values:
a = 0.250 N / 0.160 kg

Step 2: Calculate the velocity of the puck.
Calculate the change in velocity from t = 2.00 s to t = 7.00 s.
Using the formula v = u + at (where u= initial velocity, a=acceleration, t= time elapsed),
we'll calculate v = (0 m/s + a * 5.00 s) + a * (7-5) s

Step 3: Calculate the position of the puck.
To determine the change in position from t = 2.00 s to t = 7.00 s, we'll use the formula X = ut + 0.5at^2,
where u is the initial velocity of the puck, t is the time, and a is the acceleration.
Calculate X = 0m/s * (7.00s-2.00s) + 0.5 * a * (7.00s-2.00s)^2

Now, you can plug in the given values and calculate the position and speed for each scenario.