The position as a function of time is $x(t) = a + bt^2$.

Instantaneous velocity is the derivative of position with respect to time:

$v(t) = \frac{dx}{dt} = \frac{d}{dt}(a + bt^2) = 0 + 2bt = 2bt$.

Given $b = 2.5 \text{ m s}^{-2}$. So, $v(t) = 2(2.5 \text{ m s}^{-2})t = 5.0 t \text{ m s}^{-1}$.

Velocity at t = 0 s: $v(0) = 5.0 \times 0 = 0 \text{ m s}^{-1}$.

Velocity at t = 2.0 s: $v(2.0) = 5.0 \times 2.0 = 10 \text{ m s}^{-1}$.

Average velocity between t = 2.0 s and t = 4.0 s is given by $\frac{\Delta x}{\Delta t} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}$, where $t_1 = 2.0 \text{ s}$ and $t_2 = 4.0 \text{ s}$.

$x(t) = a + bt^2$. Given $a = 8.5 \text{ m}$ and $b = 2.5 \text{ m s}^{-2}$.

$x(2.0) = 8.5 + 2.5 \times (2.0)^2 = 8.5 + 2.5 \times 4.0 = 8.5 + 10.0 = 18.5 \text{ m}$.

$x(4.0) = 8.5 + 2.5 \times (4.0)^2 = 8.5 + 2.5 \times 16.0 = 8.5 + 40.0 = 48.5 \text{ m}$.

Average velocity = $\frac{x(4.0) - x(2.0)}{4.0 - 2.0} = \frac{48.5 \text{ m} - 18.5 \text{ m}}{2.0 \text{ s}} = \frac{30.0 \text{ m}}{2.0 \text{ s}} = 15 \text{ m s}^{-1}$.

The velocity at t = 0 s is 0 m s⁻¹, the velocity at t = 2.0 s is 10 m s⁻¹, and the average velocity between t = 2.0 s and t = 4.0 s is 15 m s⁻¹.

We start with the definitions of instantaneous acceleration and velocity as derivatives.

1. From the definition of acceleration, $a = \frac{dv}{dt}$. Since acceleration is constant, we can integrate this with respect to time:

$dv = a dt$

$\int_{v_0}^{v} dv = \int_{0}^{t} a dt$

$[v]_{v_0}^{v} = a \int_{0}^{t} dt$ (since a is constant)

$v - v_0 = a [t]_{0}^{t}$

$v - v_0 = a(t - 0)$

$v = v_0 + at$. (This is the first kinematic equation).

2. From the definition of velocity, $v = \frac{dx}{dt}$. We know $v = v_0 + at$. Substitute this into the definition of velocity:

$\frac{dx}{dt} = v_0 + at$

$dx = (v_0 + at) dt$

Integrate both sides with respect to time, assuming initial position $x_0$ at time $t=0$:

$\int_{x_0}^{x} dx = \int_{0}^{t} (v_0 + at) dt$

$[x]_{x_0}^{x} = \int_{0}^{t} v_0 dt + \int_{0}^{t} at dt$

$x - x_0 = v_0 \int_{0}^{t} dt + a \int_{0}^{t} t dt$ (since $v_0$ and $a$ are constant)

$x - x_0 = v_0 [t]_{0}^{t} + a [\frac{t^2}{2}]_{0}^{t}$

$x - x_0 = v_0 (t - 0) + a (\frac{t^2}{2} - \frac{0^2}{2})$

$x - x_0 = v_0 t + \frac{1}{2} at^2$. (This is the second kinematic equation, in general form).

3. To obtain the third equation ($v^2 = v_0^2 + 2a(x - x_0)$), we can use the chain rule to write acceleration $a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}$.

$a = v \frac{dv}{dx}$

$v dv = a dx$

Integrate both sides, from initial velocity $v_0$ at position $x_0$ to final velocity $v$ at position $x$:

$\int_{v_0}^{v} v dv = \int_{x_0}^{x} a dx$

$[\frac{v^2}{2}]_{v_0}^{v} = a \int_{x_0}^{x} dx$ (since a is constant)

$\frac{v^2}{2} - \frac{v_0^2}{2} = a [x]_{x_0}^{x}$

$\frac{v^2 - v_0^2}{2} = a (x - x_0)$

$v^2 - v_0^2 = 2a(x - x_0)$

$v^2 = v_0^2 + 2a(x - x_0)$. (This is the third kinematic equation, in general form).

The calculus method allows derivation of these equations and is also applicable to cases where acceleration is not constant.

Let's set up a coordinate system. Let the ground be the origin ($y=0$). Vertically upward direction is positive (+y). The ball is thrown from $y_0 = 25.0 \text{ m}$. The initial velocity is vertically upwards, so $v_0 = +20 \text{ m s}^{-1}$. The acceleration due to gravity is always downwards, so $a = -g = -10 \text{ m s}^{-2}$. This acceleration is constant during the motion.

(a) How high will the ball rise? The ball rises until its instantaneous velocity becomes zero at the highest point ($v = 0$). We need to find the position $y$ at which this occurs.

We can use the kinematic equation $v^2 = v_0^2 + 2a(y - y_0)$.

$0^2 = (20 \text{ m s}^{-1})^2 + 2(-10 \text{ m s}^{-2})(y - 25.0 \text{ m})$.

$0 = 400 \text{ m}^2 \text{ s}^{-2} - 20 \text{ m s}^{-2} (y - 25.0 \text{ m})$.

$20 \text{ m s}^{-2} (y - 25.0 \text{ m}) = 400 \text{ m}^2 \text{ s}^{-2}$.

$y - 25.0 \text{ m} = \frac{400 \text{ m}^2 \text{ s}^{-2}}{20 \text{ m s}^{-2}} = 20 \text{ m}$.

$y = 25.0 \text{ m} + 20 \text{ m} = 45.0 \text{ m}$.

The ball will rise to a height of 45.0 m from the ground.

(b) How long will it be before the ball hits the ground? The ball hits the ground when its position $y = 0 \text{ m}$. We need to find the time $t$ when $y=0$.

We can use the kinematic equation $y = y_0 + v_0 t + \frac{1}{2} at^2$.

Initial position $y_0 = 25.0 \text{ m}$, initial velocity $v_0 = +20 \text{ m s}^{-1}$, acceleration $a = -10 \text{ m s}^{-2}$. Final position $y = 0 \text{ m}$.

$0 = 25.0 + 20t + \frac{1}{2}(-10)t^2$.

$0 = 25.0 + 20t - 5t^2$.

Rearranging into a standard quadratic form $at^2 + bt + c = 0$:

$5t^2 - 20t - 25 = 0$.

Divide by 5: $t^2 - 4t - 5 = 0$.

Factor the quadratic equation: $(t - 5)(t + 1) = 0$.

Possible solutions for t are $t = 5$ s and $t = -1$ s.

Since time must be positive, the relevant solution is $t = 5$ s.

The ball will hit the ground after 5 s.

Note: Using the full path from initial to final position is simpler than splitting the motion into upward and downward segments when acceleration is constant throughout.

Free fall refers to the motion of an object solely under the influence of gravity, with air resistance being negligible. Near the Earth's surface, the acceleration due to gravity ($g$) is approximately constant and directed downwards. We usually take $g \approx 9.8 \text{ m s}^{-2}$. Free fall is a case of uniformly accelerated motion.

Let's choose the vertically upward direction as the positive y-axis, with the origin ($y=0$) at the point where the object is released (or some reference point). Since gravity acts downwards, the acceleration is $a = -g = -9.8 \text{ m s}^{-2}$.

If the object is released from rest, its initial velocity is $v_0 = 0$. The kinematic equations for motion starting from the origin ($y_0=0$) at $t=0$ with initial velocity $v_0=0$ and constant acceleration $a=-g$ become:

• Velocity as a function of time: $v = v_0 + at \implies v = 0 + (-g)t = -gt$.
• Position as a function of time: $y = y_0 + v_0 t + \frac{1}{2} at^2 \implies y = 0 + 0t + \frac{1}{2}(-g)t^2 = -\frac{1}{2}gt^2$.
• Velocity as a function of position: $v^2 = v_0^2 + 2a(y - y_0) \implies v^2 = 0^2 + 2(-g)(y - 0) = -2gy$.

These equations describe how the velocity increases in the negative direction, the position becomes more negative (further below the origin), and how velocity relates to the distance fallen. The graphs for free fall (starting from rest) are:

• Acceleration-time (a-t): A horizontal line at $a = -g$.
• Velocity-time (v-t): A straight line with a negative slope (-g) passing through the origin.
• Position-time (y-t): A downward-opening parabola starting at the origin ($y=0$, $t=0$).

If the object is thrown upwards, the initial velocity $v_0$ would be positive, and the subsequent motion is still under constant downward acceleration $-g$. The equations would be $v = v_0 - gt$, $y = y_0 + v_0 t - \frac{1}{2} gt^2$, and $v^2 = v_0^2 - 2g(y - y_0)$.

Consider an object falling from rest ($v_0 = 0$) under free fall with constant downward acceleration $g$. Let's choose the origin ($y=0$) at the starting point and the downward direction as positive. Then $y_0 = 0$ and $a = +g$.

The position of the object at time $t$ is given by $y = y_0 + v_0 t + \frac{1}{2} at^2$. With $y_0=0$ and $v_0=0$ and $a=g$, this simplifies to $y(t) = \frac{1}{2}gt^2$.

Let's consider equal intervals of time $\Delta t$. Let the time intervals be $t_1 = \Delta t$, $t_2 = 2\Delta t$, $t_3 = 3\Delta t$, etc. The positions of the object at these times are:

• At $t_1 = \Delta t$: $y_1 = \frac{1}{2}g(\Delta t)^2$.
• At $t_2 = 2\Delta t$: $y_2 = \frac{1}{2}g(2\Delta t)^2 = \frac{1}{2}g(4(\Delta t)^2) = 4 (\frac{1}{2}g(\Delta t)^2) = 4y_1$.
• At $t_3 = 3\Delta t$: $y_3 = \frac{1}{2}g(3\Delta t)^2 = \frac{1}{2}g(9(\Delta t)^2) = 9 (\frac{1}{2}g(\Delta t)^2) = 9y_1$.
• At $t_n = n\Delta t$: $y_n = \frac{1}{2}g(n\Delta t)^2 = n^2 (\frac{1}{2}g(\Delta t)^2) = n^2 y_1$.

The distance traversed during the first time interval (from $t=0$ to $t=\Delta t$) is $d_1 = y_1 - y_0 = y_1 - 0 = y_1$.

The distance traversed during the second time interval (from $t=\Delta t$ to $t=2\Delta t$) is $d_2 = y_2 - y_1 = 4y_1 - y_1 = 3y_1$.

The distance traversed during the third time interval (from $t=2\Delta t$ to $t=3\Delta t$) is $d_3 = y_3 - y_2 = 9y_1 - 4y_1 = 5y_1$.

The distance traversed during the n-th time interval (from $t=(n-1)\Delta t$ to $t=n\Delta t$) is $d_n = y_n - y_{n-1} = n^2 y_1 - (n-1)^2 y_1 = [n^2 - (n^2 - 2n + 1)] y_1 = (2n - 1) y_1$.

The ratio of the distances traversed in successive equal time intervals is:

$d_1 : d_2 : d_3 : \dots : d_n$

$y_1 : 3y_1 : 5y_1 : \dots : (2n - 1)y_1$

Dividing by $y_1$ (which is the distance covered in the first interval), we get the ratio:

$1 : 3 : 5 : \dots : (2n - 1)$

This proves Galileo's law of odd numbers: the distances traversed during equal time intervals by an object falling from rest are in the ratio of successive odd numbers starting from 1.

Let the initial velocity of the vehicle be $v_0$. When brakes are applied, the acceleration is opposite to the direction of motion, causing deceleration. Let the magnitude of deceleration be $a_{brake}$, so the acceleration in the direction of motion is $-a_{brake}$. The final velocity when the vehicle stops is $v = 0$. Let the stopping distance be $d_s$.

We can use the kinematic equation $v^2 = v_0^2 + 2a(x - x_0)$. Let the initial position be $x_0 = 0$ and the final position be $x = d_s$. The acceleration is $a = -a_{brake}$.

$0^2 = v_0^2 + 2(-a_{brake})(d_s - 0)$.

$0 = v_0^2 - 2a_{brake} d_s$.

Rearranging to solve for the stopping distance $d_s$:

$2a_{brake} d_s = v_0^2$.

$d_s = \frac{v_0^2}{2a_{brake}}$.

The stopping distance is directly proportional to the square of the initial velocity and inversely proportional to the deceleration caused by braking.

This means if you double your initial speed, your stopping distance increases by a factor of four.

The ruler is dropped under free fall. Let's choose the origin ($y=0$) at the point where the ruler is released (your friend's fingers) and the downward direction as positive (+y). The ruler starts from rest, so initial velocity $v_0 = 0$. The acceleration is due to gravity in the downward direction, $a = +g = +9.8 \text{ m s}^{-2}$. The distance the ruler falls before you catch it is $d = 21.0 \text{ cm} = 0.210 \text{ m}$. The time taken for this fall is your reaction time, $t_r$.

We can use the kinematic equation for position as a function of time: $y = y_0 + v_0 t + \frac{1}{2} at^2$.

Let $y_0 = 0$. The distance fallen is $d$, so $y = d$ at time $t = t_r$.

$d = 0 + 0 \times t_r + \frac{1}{2} g t_r^2$.

$d = \frac{1}{2} g t_r^2$.

We want to find $t_r$. Rearrange the equation:

$t_r^2 = \frac{2d}{g}$.

$t_r = \sqrt{\frac{2d}{g}}$.

Substitute the given values: $d = 0.210 \text{ m}$ and $g = 9.8 \text{ m s}^{-2}$.

$t_r = \sqrt{\frac{2 \times 0.210 \text{ m}}{9.8 \text{ m s}^{-2}}} = \sqrt{\frac{0.420 \text{ m}}{9.8 \text{ m s}^{-2}}} = \sqrt{0.042857 \text{ s}^2}$.

$t_r \approx 0.207 \text{ s}$.

Your reaction time is approximately 0.207 s.