Question

A train travels 8.81 m/s in a -51.0° direction. The train accelerated for 2.23 s, changing its velocity to 9.66 m/s in a 37.0° direction. What is (delta)x? ​

Answer 1

To find the change in position (Δx) of the train, first calculate the x-components of the initial and final velocities using the cosine of their respective angles. Then compute the average x-velocity and multiply it by the time of acceleration to obtain Δx.

Step-by-step explanation:

To calculate the change in position (Δx) of the train, we need to analyze the train's movement during the acceleration period. First, we need to break down the initial and final velocities into their horizontal x-components, using cosine as the direction is given in degrees relative to the positive x-axis.

The initial x-component of velocity (vxi) is given by:

vxi = vi × cos(θi)

For the initial velocity vi = 8.81 m/s and θi = -51.0°, the x-component is:

vxi = 8.81 × cos(-51.0°)

Similarly, the final x-component of velocity (vxf) is calculated using the final velocity vf = 9.66 m/s and θf = 37.0°, as follows:

vxf = 9.66 × cos(37.0°)

With these components, we can determine the average x-velocity (αvx) during the acceleration:

αvx = (vxi + vxf) / 2

Now, we'll use the time of acceleration (t = 2.23 s) to calculate the change in position (Δx):

Δx = αvx × t

We need to calculate the values of vxi and vxf, find αvx, and then calculate Δx using the above equation.

Answer 2

the answer is 14.77 for ∆x

the answer for ∆y is 1.16