Calculus Assignment: Displacement, Acceleration, Kinetic Energy, Charge and Volume Calculations

CALCULUS ASSIGNMENT
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Part A
(a) An expression of the displacement X of the mass is given as the integral of the
velocity.
X=ʃv; X=∫
0
10
0.8 sin ( 8 t )
(b) The displacement of the mass during the interval of the time from 0 to 3π/16 seconds.
X= ∫
0
3 π /16
0.8 sin ( 8 t )
0.8[-cos8]03π/16
=149cm
(c) An expression of the acceleration is given as a derivative of the velocity Stenger 2012
Acceleration, a= dv
dt
(d) The acceleration of the mass will be calculated as follows
=dv/dt(0.8sin8t)
The derivative of sine is cosine
=8*cos 4.712;
=-0.0031118ms-2 the negative value implies deceleration of the mass.
(e)The coordinates of the maxima and minima; at the maximum point, the derivative of x
with respect to t is zero. This implies that dx/dt=0 but dx/dv=Velocity.
0=0.8sin(8t); t=0 and therefore the cooirdinate3s of the points is at (0,1) for maximum and
(0,0)

F) Kinetic energy is given by 1/2mv2 but V=0.8sin(8t), this follows that the kinetic energy
can be expressed as K. E=1/2m[0.8sin(8t)]2
(g) The co ordinate of the maxima and minima as per the equation of the K.E are obtained by
equating the derivative of the entire equation to erode/dt=0 and this gives the coordinates as
(0,0) for the minimum and (1,2.3) for the maximum. The coordinates of the point of inflexion
is (1,2.3).
(h) The coordinates of the points of the maximum and minimum point for the Gravitational
potential Energy will be obtained by getting the derivative of the equation GPE=mg(0.3+X)
but X is the integral of the equation 0.8sin (8t) Trefethen and Weideman 2014 This means
that the required equation becomes 0=mg(o.3+0.8sin8t). The derivative of this function gives
the coordinates as (0.53.0.2) for the maximum and (0,0.2) as the minimum values. The
turning point is (0.50,0.1).
Part B
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
time
velocity
Using the trapezoidal rule, the area can be estimated as follows.1/2(0.44+0.6) x (0.8-0.6)
+1/2(0.44+0.54) x (0.8—0.4) =0.00289 square units Wu et al 2012