Question

**THERE MUST BE A GRAPHICAL METHOD FOR CALCULATING THE GRADIENT**

Question:
The equation for a distance, s (m), travelled in time t (s) by an object starting with an initial velocity u (ms^-1) and uniform acceleration a (ms^-2) is:

S= ut + 1 / 2 at^2

A) plot a graph showing a graphical method for calculating gradient of distance (s) vs time (t) for the first 10 seconds of motion if u = 10ms^-1 and a= 5ms^-2?

B) determine the gradient of the graph at t= 2s and t= 6s?

C) differentiate the equation to find the functions for:
i) velocity (v= ds/dt)
ii) acceleration (a= dv/dt = d^2 s / dt^2)

D) use your result from part c to calculate the velocity at t= 2s and t= 6s?

E) compare results from parts b and d?

Answer 1

A)$S=10t+5t^2/2$

$S=(5/2)[(t+2)^2-10]$

At $t=10$ ,

From the graph,$x- intercept=4.41$

$y-intercept=-265$

$dS/dt=265/4.41=60.1 m/s$

B)At $t=2s$

From the graph,

$x- intercept=1.25$

$y- intercept=-25$

$dS/dt=20m/s$

At $t=4s$

$x- intercept=1.83$

$y- intercept=-55$

$dS/dt=55/1.83=30.054m/s$

C) $S=ut+(1/2)at^2$

$dS/dt=u+at$

$v=u+at$

$dv/dt=a$

$acceleration=a$

D)$v=u+at$

At $t=2s$

$v=10+5×2=20m/s$

At $t=4s$

$v=10+5×4=30m/s$

E)The answers obtained in Part B and Part D are approximately equal.