1. Fundamental Relationships The total acceleration $\vec{a}$ of the car is the vector sum of its tangential component $\vec{a}_t$ and normal (centripetal) component $\vec{a}_n$.

$$ \vec{a} = \vec{a}_t + \vec{a}_n $$

For uniform deceleration, the magnitude of the tangential acceleration $a_{t,mag} = |\vec{a}_t|$ is constant. The vector $\vec{a}_t$ is anti-parallel to the velocity vector $\vec{v}$. The vector $\vec{a}_n$ is perpendicular to $\vec{v}$. The relationship between the angle $\theta$ (between $\vec{a}$ and $\vec{v}$) and the acceleration components is:

$$ a_n = -a_{t,mag} \tan(\theta) $$

The speed $v$ at time $t$ and the normal acceleration $a_n$ are given by:

$$ v(t) = v_0 - a_{t,mag} t $$ $$ a_n = \frac{v(t)^2}{R} $$

where $R$ is the radius of the track.

2. Derivation of Key Parameters We apply these relations at the two specified times, $t_1$ and $t_{12} = t_1 + t_2$. At time $t_1$:

$$ a_{n1} = \frac{v_1^2}{R} = \frac{(v_0 - a_{t,mag} t_1)^2}{R} = -a_{t,mag} \tan(\theta_1) \quad (1) $$

At time $t_{12}$:

$$ a_{n2} = \frac{v_2^2}{R} = \frac{(v_0 - a_{t,mag} t_{12})^2}{R} = -a_{t,mag} \tan(\theta_2) \quad (2) $$

To find $a_{t,mag}$, we eliminate $R$ by taking the ratio of equation (1) and (2):

$$ \frac{(v_0 - a_{t,mag} t_1)^2}{(v_0 - a_{t,mag} t_{12})^2} = \frac{\tan(\theta_1)}{\tan(\theta_2)} $$

Taking the square root and defining $k = \sqrt{\frac{\tan(\theta_1)}{\tan(\theta_2)}}$:

$$ \frac{v_0 - a_{t,mag} t_1}{v_0 - a_{t,mag} t_{12}} = k $$

Solving for $a_{t,mag}$:

$$ v_0 - a_{t,mag} t_1 = k(v_0 - a_{t,mag} t_{12}) $$ $$ a_{t,mag} (k t_{12} - t_1) = v_0 (k - 1) $$ $$ a_{t,mag} = \frac{v_0 (k - 1)}{k t_{12} - t_1} $$

3. Calculations First, we calculate the constant $k$:

$t_1 = 5$ s, $t_{12} = 5+3=8$ s, $\theta_1 = 135^\circ$, $\theta_2 = 150^\circ$. $$ k = \sqrt{\frac{\tan(135^\circ)}{\tan(150^\circ)}} = \sqrt{\frac{-1}{-1/\sqrt{3}}} = \sqrt{\sqrt{3}} = 3^{1/4} \approx 1.316 $$

[Q2] Tangential Acceleration, $a_{t,mag}$ Substitute the values into the derived expression for $a_{t,mag}$:

$$ a_{t,mag} = \frac{7.0(3^{1/4} - 1)}{3^{1/4}(8) - 5} = \frac{7.0(1.316 - 1)}{1.316(8) - 5} = \frac{7.0(0.316)}{10.528 - 5} \approx 0.40 \text{ m/s}^2 $$

[Q3] Normal Accelerations, $a_{n1}$ and $a_{n2}$ Using the value of $a_{t,mag}$:

$$ a_{n1} = -a_{t,mag} \tan(\theta_1) = -(0.40 \text{ m/s}^2) \tan(135^\circ) = -(0.40)(-1) = 0.40 \text{ m/s}^2 $$ $$ a_{n2} = -a_{t,mag} \tan(\theta_2) = -(0.40 \text{ m/s}^2) \tan(150^\circ) = -(0.40)(-1/\sqrt{3}) \approx 0.23 \text{ m/s}^2 $$

[Q1] Radius of the Track, $R$ First, find the speed at $t_1$:

$$ v_1 = v_0 - a_{t,mag} t_1 = 7.0 \text{ m/s} - (0.40 \text{ m/s}^2)(5 \text{ s}) = 5.0 \text{ m/s} $$

Now, use the definition of normal acceleration $a_{n1} = v_1^2 / R$:

$$ R = \frac{v_1^2}{a_{n1}} = \frac{(5.0 \text{ m/s})^2}{0.40 \text{ m/s}^2} = \frac{25}{0.40} = 62.5 \text{ m} $$

Rounding to two significant figures, $R \approx 63$ m.