Question

The distance, x, covered by a particle in time, t, is given as x=a +bc+ct^2 +dt^3

Answer 1

Answer:

$a$ has units of distance

$b$  has units of distance over time

$c$  has units of distance over $time^2$

$d$ has units of distance over $time^3$

Explanation:

Since the expression for the distance is:

$x = a+b\,t+c\,t^2+d\,t^3$

then:

$a$ has units of distance

$b$  has units of distance over time

$c$  has units of distance over $time^2$

$d$ has units of distance over $time^3$

because we are supposed to be able to add all of the terms and get a distance.  So the products on each term that contains factors of time (t) should be cancelling those time units with units in the denominator of the multiplicative constant s that accompany them.