Answer to Question #138324 in Calculus for lilililik,

Question

The equation for a distance s(m), travelled in time t(s) by an object is given by:

s=2t2-3t+5
The tasks are to:

a) Plot a graph of distance (s) vs time (t) for the first 5s of motion
b) Determine the gradient of the graph at t=2s and t=4s.
c) Identify the position of any turning points and whether they are maxima, minima or points of inflexion.
d) Differentiate the equation to find the functions for
i) Velocity (v=dsdt)
ii) Acceleration (a=dvdt=d2sdt2)

e) Use your result from part d to calculate the velocity at t=2s and t=4s.
f) Calculate the turning points of the function using differential calculus and show which are maxima, minima or points of inflexion by using the second derivative.
g) Compare your results for part b and part e.

Answer 1

y = 2t² - 3t + 5

a)

b)

Gradient at t = 2s

= $[\frac{ds}{dt} ]_{t=2} = [4t-3]_{t=2}=5$

Gradient at t = 4s

= $[\frac{ds}{dt} ]_{t=4} = [4t-3]_{t=4}=13$

c) There is a turning point at A(3/4, 31/8) which is a point of minima

d) velocity, v = $\frac{ds}{dt} = (4t-3) m/s$

Acceleration, a = $\frac{dv}{dt} = 4$ m/s²

e) velocity at t = 2

= $[v ]_{t=2}= [4t-3]_{t=2}=5$ m/s

Velocity at t = 4s

= $[v ]_{t=4}= [4t-3]_{t=4}=13$ m/s

f)

s = 2t² - 3t + 5

$\frac{ds}{dt} = 0 => 4t-3 = 0$

=> t = 3/4

So the function has turning point at t = 3/4

$\frac{d²s}{dt²} = 4 > 0$

As second derivative is positive at the turning point, it is a point of minima.

g) Results of part b and part e are equal . So gradient represents the velocity.