Imagine that if you step hard a certain car\'s accelerator at t=0, its forward a
ID: 3900140 • Letter: I
Question
Imagine that if you step hard a certain car's accelerator at t=0, its forward acceleration is given by
ax(t)=a0sinwt for 0 ? wt ? pi and becomes zero afterward, where a0 = 5.0 m/s^2 and q is a constant with a value of pi/5 when expressed in the approximate SI units. Note that the car's acceleration is initially zero when t=0 but increases to a maximum of a0 as the engine reaches its maximally efficient speed, but then decreases as drag and friction forces begin to oppose the car's motion significantly. When the car reaches its cruising speed at a time t such that wt=pi, the car's acceleration becomes zero. (This is not likely to be the actual acceleration function for any realistic car, but we can use it as a simple model.)
a) What are the SI units of q?
b) Assume that the car starts from rest at x=0.
What are the car's x-velocity and x-position at time t=5.0s? (Hint: the derivative of sinwt is wcoswt, and the derivative of coswt is -sinwt. What, therefore, and are the antiderivatives of sinwt and coswt?)
Explanation / Answer
a)
SI units of q will be s^-1.
b)
a = a0*sin wt
a = dv/dt
dv / dt = a0*sin wt
Integrating it with time
v = a0*(-cos wt) / w + C1
Now at t = 0 we are given that v = 0.
So, 0 = a0*(-cos 0) / w + C1
C1 = a0 / w
Thus, v = (a0 / w) * (1 - cos wt)..........................1
Now v = dx / dt
dx / dt = (a0 / w) * (1 - cos wt)
Integrating it with time
x = a0/w * (t - (sin wt) /w) + C2
Now at t = 0 we are given that x = 0.
So, 0 = a0/w * (t - (sin 0) /w) + C2
So, C2 = -a0*t / w
Therefore,
x = a0/w * (t - (sin wt) /w) - a0*t / w
x = (a0 / w) [t - (sin wt) / w - t]
x = -a0*(sin wt) / w^2................................2
At t = 5 s, from eqns 1 and 2 we get
putting w = pi/5, a0 = 5
v = 5/(pi/5) * [1 - cos pi/5 *5]
v = 25/pi *2
v = 50 / pi
x = -5 / (pi/5)^2 *sin (pi/5 *5)
x = 0