Question

The position of a particle moving along the x axis depends on the time according to the equation x = ct² - bt³, where x is in meters and t in seconds.
What are the units of (a) constant c and (b) constant b? Let their numerical values be 3.0 and 2.0, respectively.
(c) At what time does the particle reach its maximum positive x position? From t = 0.0 s to t = 4.0 s,
(d) what distance does the particle move and
(e) what is its displacement?
Find its velocity at times (f) 1.0 s, (g) 2.0 s, (h) 3.0 s, and (i) 4.0 s.
Find its acceleration at times (j) 1.0 s, (k) 2.0 s, (l) 3.0 s, and (m) 4.0 s.

Answer 1

Answer:

(a) $\frac{m}{s^2}$

(b) $\frac{m}{s^3}$

(c) 1 s

(d) 20 m

(e) 1 m

(f) $0\frac{m}{s}$

(g) $-12\frac{m}{s}$

(h) $-36\frac{m}{s}$

(i) $-72\frac{m}{s}$

(j) $-6\frac{m}{s^2}$

(k) $-18\frac{m}{s^2}$

(l) $-30\frac{m}{s^2}$

(m) $-42\frac{m}{s^2}$

Explanation:

Since x is measured in meters and t in seconds, constants a and b must have units that gives meters when multiplied by square and cubic seconds respectivly, so that would mean $\frac{m}{s^2}$ for a and $\frac{m}{s^3}$ for b.

We can get the velocity v equation by deriving the position with respect to t, which gives:

$v=6*t-6*t^2$

And the acceleration a equation by deriving again:

$a=6-12*t$

Now for getting the maximun position between 0 and 4, we must find to points where the positions first derivate is equal to cero and evaluate those points. That is v=0, which gives

$6*t-6*t^2=0\\6t*(1-t)=0\\t=0 or t=1$

For t = 0, x = 0 so the maximun position is archieved at 1 second, which gives x = 1 meter.

For obtaining it's displacement r, we can integrate the velocity from 0 seconds to 4 seconds, which gives the mean value of the position in that interval:

$r=\frac{1}{4}*(2*4^3-3*4^2)m\\r= 20m$

For the remaining questions, we just replace the values of t on the respective equations.