Answer:
[tex]1.68 \frac{m}{s^2}[/tex]
Explanation:
Please find the image for the question as attached file.
Solution -
Given -
First of all we will calculate the velocity at point C,
As per newton's third law of motion-
[tex]V_C^2 = V_A^2 + 2 a_t (S_C - S_A)\\[/tex]
Substituting the given values in above equation, we get -
[tex]V_C^2 = 20^2 + 2*0.5*(100-0)\\V_C = 22.361 \frac{m}{s}[/tex]
Now we will determine the radius of curvature for the curve shown in the attached image
[tex]Y = 16 - \frac{1}{625} X^2\\[/tex]
Differentiating on both the sides, we get -
[tex]\frac{dy}{dx} = -3.2 (10^-3) X\\\frac{d^2y}{d^2x} = -3.2 (10^-3)\\Curve = \frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}} }{\frac{d^2y}{d^2x}} \\Curve = 312.5[/tex]meter
Acceleration on curved path
[tex]a = \frac{V_C^2}{Curve} \\a = \frac{22.361^2}{312.5} \\a= 1.60 \frac{m}{s^2}[/tex]
Final acceleration
[tex]a_f = \sqrt{0.5^2 + 1.6^2} \\a_f = 1.68\frac{m}{s^2}[/tex]
