Question

A slowly moving flat car is 12.0 m long passing apoint at straight road at 10 kph. A boy beside the road near to that point tosses rocks onto the moving flat car at the rate of one per second, (a) If the first rock just hits the front edge of the car, how many rocks will fall on to that car ? (b) How many rocks will fall onto that car if the car begins to accelerate at 0.5 m//s^(2), just as the first rock hits the car ?

Answer 1

(a) 5 rocks, (b) 4 rocks

Answer 2

The problem describes a scenario where a boy tosses rocks onto a moving flat car. We need to determine the number of rocks that fall onto the car under two different conditions: (a) constant velocity and (b) constant acceleration.

Let's define the coordinate system: Let the origin $x=0$ be the fixed point on the road from where the boy tosses the rocks. The rocks are tossed at a rate of one per second, so they are tossed at $t=0, 1, 2, 3, \dots$ seconds. The problem states that the first rock (tossed at $t=0$) just hits the front edge of the car. This means at $t=0$, the front edge of the car is at $x=0$. The length of the car is $L = 12.0 \text{ m}$. The initial speed of the car is $v_0 = 10 \text{ kph}$. We need to convert $v_0$ to m/s: $v_0 = 10 \text{ kph} = 10 \times \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{100}{36} \text{ m/s} = \frac{25}{9} \text{ m/s}$.

Let $x_f(t)$ be the position of the front edge of the car at time $t$. Let $x_r(t)$ be the position of the rear edge of the car at time $t$. Since the front edge is at $x=0$ at $t=0$, the rear edge is at $x=-L$ at $t=0$. The car moves in the positive x-direction. For a rock tossed at time $t_n$ to fall on the car, the car must be covering the point $x=0$ at that instant. This means the point $x=0$ must be between the rear edge and the front edge of the car (inclusive). So, $x_r(t_n) \le 0 \le x_f(t_n)$.

Part (a): Constant velocity The car moves with constant velocity $v_0$. The position of the front edge at time $t$ is $x_f(t) = v_0 t$. The position of the rear edge at time $t$ is $x_r(t) = v_0 t - L$.

For a rock tossed at time $t_n$ to fall on the car, we must satisfy:

1. $x_f(t_n) \ge 0 \implies v_0 t_n \ge 0$. This is always true for $t_n \ge 0$ since $v_0 > 0$.
2. $x_r(t_n) \le 0 \implies v_0 t_n - L \le 0 \implies v_0 t_n \le L \implies t_n \le \frac{L}{v_0}$.

Let's calculate the maximum time $t_{max}$ for a rock to fall on the car: $t_{max} = \frac{L}{v_0} = \frac{12.0 \text{ m}}{25/9 \text{ m/s}} = \frac{12 \times 9}{25} \text{ s} = \frac{108}{25} \text{ s} = 4.32 \text{ s}$.

The rocks are tossed at $t=0, 1, 2, 3, 4, 5, \dots$ seconds. For a rock to fall on the car, its toss time $t_n$ must be less than or equal to $4.32 \text{ s}$. The integer values of $t_n$ that satisfy this condition are $t_n = 0, 1, 2, 3, 4$. These correspond to 5 rocks.

Part (b): Accelerating car The car begins to accelerate at $a = 0.5 \text{ m/s}^2$ just as the first rock hits the car ($t=0$). The position of the front edge at time $t$ is $x_f(t) = v_0 t + \frac{1}{2} a t^2$. The position of the rear edge at time $t$ is $x_r(t) = v_0 t + \frac{1}{2} a t^2 - L$.

For a rock tossed at time $t_n$ to fall on the car, we must satisfy:

1. $x_f(t_n) \ge 0 \implies v_0 t_n + \frac{1}{2} a t_n^2 \ge 0$. This is always true for $t_n \ge 0$ since $v_0 > 0$ and $a > 0$.
2. $x_r(t_n) \le 0 \implies v_0 t_n + \frac{1}{2} a t_n^2 - L \le 0$.

Substitute the given values: $v_0 = \frac{25}{9} \text{ m/s}$, $a = 0.5 \text{ m/s}^2$, $L = 12.0 \text{ m}$. $\frac{25}{9} t_n + \frac{1}{2} (0.5) t_n^2 - 12 \le 0$ $\frac{25}{9} t_n + 0.25 t_n^2 - 12 \le 0$ $0.25 t_n^2 + \frac{25}{9} t_n - 12 \le 0$. To find the roots of the quadratic equation $0.25 t^2 + \frac{25}{9} t - 12 = 0$, we can multiply by 4 to simplify coefficients: $t^2 + \frac{100}{9} t - 48 = 0$.

Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $t = \frac{-\frac{100}{9} \pm \sqrt{(\frac{100}{9})^2 - 4(1)(-48)}}{2(1)}$ $t = \frac{-\frac{100}{9} \pm \sqrt{\frac{10000}{81} + 192}}{2}$ $t = \frac{-\frac{100}{9} \pm \sqrt{\frac{10000 + 192 \times 81}{81}}}{2}$ $192 \times 81 = 15552$. $t = \frac{-\frac{100}{9} \pm \sqrt{\frac{10000 + 15552}{81}}}{2}$ $t = \frac{-\frac{100}{9} \pm \sqrt{\frac{25552}{81}}}{2}$ $t = \frac{-\frac{100}{9} \pm \frac{\sqrt{25552}}{9}}{2}$ $\sqrt{25552} \approx 159.85$. $t = \frac{-100 \pm 159.85}{18}$.

We are interested in positive time, so we take the positive root: $t_{max} = \frac{-100 + 159.85}{18} = \frac{59.85}{18} \approx 3.325 \text{ s}$.

The condition for a rock to fall on the car is $t_n \le 3.325 \text{ s}$. The integer values of $t_n$ that satisfy this condition are $t_n = 0, 1, 2, 3$. These correspond to 4 rocks.