Answer to Question #161404 in Calculus for Tarunjot Singh Bhatia

Question

1 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 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 = 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 𝑑𝑡2) 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

a) Graph of distance s = 10t + 5/2 t²

b) AD is a tangent line at t=2. The coordinates of the points: A(2, 30), D(5, 90).

The gradient is (90 - 30) / (5 - 2)= 20.

BC is a tangent line at t=6. The coordinates of the points: B(6, 150), C(3, 30).

The gradient is (150 - 30) / (6 - 3)= 40.

c)$v =\frac{ds}{dt} = \frac{d}{dt}(10t + \frac{5}{2} t^2)=5t+10$

$a =\frac{d^2s}{dt^2} = \frac{d^2}{dt^2}(10t + \frac{5}{2} t^2)=5$

d) $v(2) = 5\cdot2 + 10 = 20$

$v(6) = 5\cdot 6 + 10 = 40$

e) In both cases (t=2 and t=6) the values of v(t) match with the gradients of the tangent lines at these points.