An object is at rest at (1 ,2 ,8 )(1,2,8) and constantly accelerates at a rate of 1 m/s^21ms2 as it moves to point B. If point B is at (3 ,7 ,4 )(3,7,4), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

2 Answers
Jun 9, 2017

It will take 3.66sec.

Explanation:

The distance is sqrt((1-3)^2+(2-7)^2+(8-4)^2)(13)2+(27)2+(84)2=√45
Distance between the two points is √45 m
Solving by equation,
√45=(1/2)×1×t^245=(12)×1×t2
t=3.66t=3.66

Jun 9, 2017

t = 3.66t=3.66 "s"s

Explanation:

We're asked to find the time tt it takes the object to travel from two coordinate points with a constant acceleration of 1 "m"/("s"^2)1ms2.

We can use the equation

Deltax = v_(0x)t + 1/2a_xt^2

to find this time.

The quantity Deltax is the total displacement, which is

Deltax = sqrt((3"m"-1"m")^2 + (7"m"-2"m")^2 + (4"m"-8"m")^2)

=color(red)(6.71 color(red)("m"

The intital velocity v_(0x) is 0, because it started from rest.

Plugging in known variables, we have

color(red)(6.71 color(red)("m") = (0)t + 1/2(1"m"/("s"^2))t^2

t = sqrt((6.71"m")/(1/2"m"/("s"^2))) = color(blue)(3.66 color(blue)("s"