Answer to Question #301372 in Calculus for Neeraj

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+12𝑎𝑡2 The tasks are to: a) Plot a graph of distance (s) vs time (t) for the first 10s of motion if 𝑢=10𝑚𝑠−1 and 𝑎=4𝑚𝑠−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=𝑑𝑠𝑑𝑡) ii) Acceleration (𝑎=𝑑𝑣𝑑𝑡=𝑑2𝑠𝑑𝑡2) d) Use your result from part c to calculate the velocity at 𝑡=2𝑠 and 𝑡=6𝑠. e) Compare your results for part b and part d.

Answer 1

$$s(t)=10t+\dfrac{4t^2}{2}, m$$

a)

$$s(t)=10t+2t^2, 0\le t\le10$$

b)

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

$t=2 , s(2)=28$

$$grad \ s|_{t=2}=\dfrac{10(2+\Delta t)+2(2+\Delta t)^2-28}{\Delta t}$$

$$=18+2\Delta t\to 18( m/s)$$

$t=6 , s(6)=132$

$$grad \ s|_{t=6}=\dfrac{10(6+\Delta t)+2(6+\Delta t)^2-132}{\Delta t}$$

$$=34+2\Delta t\to 34( m/s)$$

c)

i)

$$v(t)=\dfrac{ds}{dt}=10+4t, m/s$$

ii)

$$a(t)=\dfrac{dv}{dt}=\dfrac{d^2s}{dt^2}=4\ m/s^2$$

d)

$$v(2)=10+4(2)=18(m/s)$$

$$v(6)=10+4(6)=34(m/s)$$

e) The results are the same.