Question

The car passes point A with a speed of 25 m/s after which its speed is defined by v = (25 - 0.15 s) m/s. Determine the magnitude of the car's acceleration when it reaches point B, where s = 51.5 m.

Answer 1

2.75 m/s^2

Answer 2

The car's motion involves both a change in speed (tangential acceleration) and a change in direction (normal acceleration). The total acceleration is the vector sum of these two components.

1. Calculate Tangential Acceleration ($a_t$): The speed of the car is given by $v = (25 - 0.15 s)$ m/s. Tangential acceleration is the rate of change of speed, $a_t = \frac{dv}{dt}$. Since $v$ is a function of $s$, we use the chain rule: $a_t = \frac{dv}{ds} \frac{ds}{dt} = \frac{dv}{ds} v$.

First, find $\frac{dv}{ds}$: $v = 25 - 0.15 s$ $\frac{dv}{ds} = -0.15$ s$^{-1}$.

Next, calculate the speed $v$ at point B, where $s = 51.5$ m: $v_B = 25 - 0.15 (51.5)$ $v_B = 25 - 7.725$ $v_B = 17.275$ m/s.

Now, calculate the tangential acceleration $a_t$ at point B: $a_t = (-0.15) \times (17.275)$ $a_t = -2.59125$ m/s$^2$. The negative sign indicates deceleration. The magnitude of tangential acceleration is $|a_t| = 2.59125$ m/s$^2$.

2. Calculate Normal Acceleration ($a_n$): Normal acceleration is given by $a_n = \frac{v^2}{\rho}$, where $\rho$ is the radius of curvature of the path at point B. The path is defined by the equation $y = 16 - \frac{1}{625}x^2$. To find the radius of curvature $\rho$, we need the first and second derivatives of $y$ with respect to $x$: $y' = \frac{dy}{dx} = -\frac{2}{625}x$ $y'' = \frac{d^2y}{dx^2} = -\frac{2}{625}$

The formula for the radius of curvature is $\rho = \frac{[1 + (y')^2]^{3/2}}{|y''|}$.

We need to find the x-coordinate of point B corresponding to an arc length $s = 51.5$ m from point A (the vertex at $x=0$). The arc length $s$ from $x=0$ to $x$ is given by $s = \int_0^x \sqrt{1 + (y')^2} dx$. $s = \int_0^x \sqrt{1 + \left(-\frac{2t}{625}\right)^2} dt = \int_0^x \sqrt{1 + \frac{4t^2}{625^2}} dt$. This integral is difficult to solve analytically for $x$. However, for small values of $y'$, we can use the approximation $\sqrt{1+u} \approx 1 + \frac{1}{2}u$. So, $\sqrt{1 + \frac{4t^2}{625^2}} \approx 1 + \frac{1}{2}\left(\frac{4t^2}{625^2}\right) = 1 + \frac{2t^2}{625^2}$. Using this approximation: $s \approx \int_0^x \left(1 + \frac{2t^2}{625^2}\right) dt = \left[t + \frac{2t^3}{3 \times 625^2}\right]_0^x = x + \frac{2x^3}{3 \times 625^2}$. Given $s = 51.5$ m: $51.5 = x + \frac{2x^3}{3 \times (625)^2}$ $51.5 = x + \frac{2x^3}{3 \times 390625} = x + \frac{2x^3}{1171875}$. Let's find $x$ by trial and error or by checking close values: If $x=51.25$: $51.25 + \frac{2(51.25)^3}{1171875} = 51.25 + \frac{2(134673.828)}{1171875} = 51.25 + \frac{269347.656}{1171875} \approx 51.25 + 0.2298 \approx 51.4798$. This is very close to $51.5$. So we can use $x \approx 51.25$ m as the x-coordinate for point B.

Now, calculate $y'$ at $x = 51.25$ m: $y' = -\frac{2}{625}(51.25) = -\frac{102.5}{625} = -0.164$.

Now, calculate the radius of curvature $\rho$ at point B: $\rho = \frac{[1 + (-0.164)^2]^{3/2}}{|-\frac{2}{625}|} = \frac{[1 + 0.026896]^{3/2}}{\frac{2}{625}}$ $\rho = \frac{(1.026896)^{3/2}}{0.0032} = \frac{1.04046}{0.0032} \approx 325.14$ m.

Finally, calculate the normal acceleration $a_n$ at point B: $a_n = \frac{v_B^2}{\rho} = \frac{(17.275)^2}{325.14}$ $a_n = \frac{298.425625}{325.14} \approx 0.9178$ m/s$^2$.

3. Calculate the Magnitude of Total Acceleration ($a$): The magnitude of the total acceleration is $a = \sqrt{a_t^2 + a_n^2}$. $a = \sqrt{(-2.59125)^2 + (0.9178)^2}$ $a = \sqrt{6.7145 + 0.8423}$ $a = \sqrt{7.5568}$ $a \approx 2.749$ m/s$^2$.

Rounding to a reasonable number of significant figures, the magnitude of the car's acceleration is approximately $2.75$ m/s$^2$.