\[
s=u t+\frac{1}{2} a t^{2}
\]
The tasks are to:
a) Plot a graph of distance (s) vs time (t) for the first $10 \mathrm{~s}$ of motion if $u=10 \mathrm{~ms}^{-1}$ and $a=5 \mathrm{~ms}^{-2}$.
b) Determine the gradient of the graph at $t=2 \mathrm{~s}$ and $t=$ $6 s$.
c) Differentiate the equation to find the functions for
i) Velocity $\left(v=\frac{d s}{d t}\right)$
ii) Acceleration $\left(a=\frac{d v}{d t}=\frac{d^{2} s}{d t^{2}}\right)$
d) Use your result from part c to calculate the velocity at $t=2 s$ and $t=6 s$.
e) Compare your results for part b and part d.

Step 1 ：a) To plot the graph of distance (s) vs time (t), we use the given equation: \(s = ut + \frac{1}{2}at^2\) with \(u = 10\) and \(a = 5\). So, \(s = 10t + \frac{1}{2}(5)t^2 = 10t + 2.5t^2\).

Step 2 ：b) To find the gradient of the graph at \(t = 2s\) and \(t = 6s\), we need to find the first derivative of the equation with respect to time: \(\frac{ds}{dt} = \frac{d}{dt}(10t + 2.5t^2) = 10 + 5t\). At \(t = 2s\), the gradient is \(10 + 5(2) = 20\). At \(t = 6s\), the gradient is \(10 + 5(6) = 40\).

Step 3 ：c) i) The velocity function is the first derivative of the distance function: \(v = \frac{ds}{dt} = 10 + 5t\).

Step 4 ：ii) The acceleration function is the first derivative of the velocity function: \(a = \frac{dv}{dt} = \frac{d}{dt}(10 + 5t) = 5\).

Step 5 ：d) To calculate the velocity at \(t = 2s\) and \(t = 6s\), we use the velocity function: \(v = 10 + 5t\). At \(t = 2s\), the velocity is \(10 + 5(2) = \boxed{20}\). At \(t = 6s\), the velocity is \(10 + 5(6) = \boxed{40}\).

Step 6 ：e) Comparing the results from part b and part d, we see that the gradient of the graph at \(t = 2s\) and \(t = 6s\) is equal to the velocity at those times, which are \(20\) and \(40\) respectively.