The distance a cart moves down a track (from rest) is proportional to the time squared. How far would a cart move down a track (from rest) in two seconds if it moves (from rest) a distance of 20 cm in one second? A) 5 cm B) 10 cm C) 20 cm D) 40 cm E) 80 cm

A car move down a track a distance of 80 cm

Letter E  80 cm

Proportional

if the distance is proportional to the time square we can express such condition writing

d = k*t²        (1)

where d is the distance t is the time and k is the constant of proportionality.

Now we must considerer that if the cart moves a distance of 20 cm in one second then by substitution in equation (1)

20 = k * (t)²        ⇒     20 = k*1   then   k = 20

The proportionality constant is K = 20

To determine how far the cart move down in 2 seconds, again by substitution in equation (1) ( but now knowing k )

d = 20*(2)²

d = 20*4

d = 80 cm

graphing motion: mastery test

209be

what is the instantaneous acceleration of

the particle at point b?

0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

20

time (seconds)

answer: 0 meters/second^2

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm