A car is traveling around a horizontal circular track with radius r = 240 m at a constant speed v = 25 m/s as shown. The angle èA = 34° above the x axis, and the angle èB = 65° below the x axis.
I found the car's acceleration to be 2.6 m/s^2 (a=v^2/r).
Now I need to know the x component of the car's acceleration when it is at point A.
I'm not sure how to go about solving this.
whats the answerrrr ?
x-component of car's acceleration:
centripetal acceleration * cos(angle)
remember that centripetal acceleration is v^2/r
To find the x-component of the car's acceleration at point A, you can use trigonometry.
The x-axis is horizontal, so the y-component of acceleration at point A (ayA) is zero since there is no vertical acceleration.
Now, let's consider the angle èA. Since the angle èA is above the x-axis, it corresponds to the y-component of acceleration, not the x-component. To find the x-component of acceleration (axA), we need to find the length of the adjacent side of the right triangle formed by the x-axis and the acceleration vector.
To do this, we can use the fact that the tangent of an angle in a right triangle is equal to the ratio of the length of the opposite side to the length of the adjacent side.
In this case, we know the value of the tangent of angle èA (tan(èA) = opposite/adjacent), and we want to find the adjacent side (axA). The opposite side is the y-component of acceleration (ayA), which we determined to be zero.
Therefore, the tangent of angle èA is equal to zero divided by axA. Solving for axA, we find that it is also zero.
Therefore, at point A, the car's acceleration has no x-component (axA = 0).
To summarize, the x-component of the car's acceleration when it is at point A is zero.