How Does an Accelerometer Help Calculate Plane Takeoff Speed?

[lmf22]

A person dangles her watch from a string while the plane takes off. She notices that the string makes a angle of 25 degrees with respect to the vertical whie the plane accelerates to take off, which takes 19 seconds. What is the takeoff speed of the plane?

How do I get started? I think the X component is (Tension of String)*cos(25) and the Y component is (Tension of String)*sin(25)?
How do I find the accerleration without knowing the mass of the watch?

 

[Doc Al]

How do I get started? I think the X component is (Tension of String)*cos(25) and the Y component is (Tension of String)*sin(25)?

Draw yourself a picture. I think you've got the components of the tension reversed. (The angle is with respect to the vertical.) Don't forget to consider all the forces on the watch.

How do I find the accerleration without knowing the mass of the watch?

By applying Newton's 2nd law for the x and y components of the forces on the watch. (You'll find that the mass drops out.) Hint: the acceleration in the y-direction is zero.

 

[M98Ranger]

To solve this problem, we can use the equation for acceleration: a = ∆v/∆t. We know that the plane accelerates from rest to its takeoff speed in 19 seconds, so we can set ∆t = 19 seconds. We also know that the angle between the string and the vertical is 25 degrees, which means the Y component of the tension in the string will be (Tension of String)*sin(25). This Y component will be equal to the force of gravity acting on the watch, which is equal to the mass of the watch (m) multiplied by the acceleration due to gravity (g). So, we can write the equation as (Tension of String)*sin(25) = m*g.

To find the acceleration of the plane, we need to find the change in velocity (∆v) in the X-direction. We can use trigonometry to find the X-component of the tension in the string, which is (Tension of String)*cos(25). This X-component will be equal to the force that causes the plane to accelerate, which is equal to the mass of the plane (M) multiplied by the acceleration of the plane (a). So, we can write the equation as (Tension of String)*cos(25) = M*a.

Now, we can use the equation for acceleration to solve for the takeoff speed of the plane. We can rearrange the equation a = ∆v/∆t to ∆v = a*∆t. Plugging in the values we know, we get ∆v = a*19 seconds. We also know that ∆v = (Tension of String)*cos(25), so we can substitute this into the equation to get (Tension of String)*cos(25) = a*19 seconds.

Next, we can substitute the value for tension in terms of mass and acceleration due to gravity (m*g) into the equation (Tension of String)*cos(25) = a*19 seconds. This gives us (m*g)*cos(25) = a*19 seconds.

Finally, we can substitute the value for tension in terms of mass and acceleration of the plane (M*a) into the equation (Tension of String)*sin(25) = m*g. This gives us (M*a)*sin(25) = m*g.

Now, we have