Question

{F} The equation for a displacement 𝑠(𝑚), at a time 𝑡(𝑠) by an object starting at a displacement of 𝑠0 (𝑚), with an initial velocity 𝑢(𝑚𝑠 −1 ) and uniform acceleration 𝑎(𝑚𝑠 −2 ) is: 𝑠 = 𝑠0 + 𝑢𝑡 + 1 2 𝑎𝑡 2 A projectile is launched from a cliff with 𝑠0 = 30 𝑚, 𝑢 = 55 𝑚𝑠 −1 and 𝑎 = −10 𝑚𝑠 −2 . The tasks are to: a) Plot a graph of distance (𝑠) vs time (𝑡) for the first 10s of motion. 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 𝑑𝑡2 ) d) Use your results from part c to calculate the velocity at 𝑡 = 2𝑠 and 𝑡 = 6𝑠. e) Compare your results for part b) and part d). f) Find the turning point of the equation for the displacement 𝑠 and using the second derivative verify whether it is a maximum, minimum or point of inflection. g) Compare your results from f) with the graph you produced in a).

Answer 1

$$s(t)=30+55t-\dfrac{10t^2}{2}, m$$

a)

$$s(t)=30+55t-5t^2, 0\le t\le10$$

b)

$$grad \ s=\dfrac{s_2-s_1}{t_2-t_1}$$

$t=2 , s(2)=120$

$t=0 , s=50$

$$grad \ s|_{t=2}=\dfrac{120-50}{2-0}=35(m/s)$$

$t=6 , s(6)=180$

$t=0 , s=210$

$$grad \ s|_{t=6}=\dfrac{180-210}{6-0}=-5(m/s)$$

c)

i)

$$v(t)=\dfrac{ds}{dt}=55-10t, m/s$$

ii)

$$a(t)=\dfrac{dv}{dt}=\dfrac{d^2s}{dt^2}=-10\ m/s^2$$

d)

$$v(2)=55-10(2)=35(m/s)$$

$$v(6)=55-10(6)=-5(m/s)$$

e) The results are the same.

f) The turning point of the equation for the displacement

$$\dfrac{ds}{dt}=0=>55-10t=0$$

$$t=5.5 s$$

$$s(5.5)=30+55(5.5)-5(5.5)^2=181.25(m)$$

Turning point is $(5.5, 181.25)$

$$\dfrac{d^2s}{dt^2}=-10<0$$

Point $(5.5, 181.25)$ is a maximum.

g) The results are the same.