Where is the particle?

Calculus Level 2

A particle is confined to moving along the $x$ -axis. The initial conditions are a velocity of 0 and a starting position of $x = 1 \text{ meter}$ .

A constant acceleration is applied (a = 4 $\text{m/sec}^2$ ) for 5 seconds. Where is the particle after 5 seconds?

The particle is moving in the direction of increasing $x$ .

3 solutions

Zach Abueg
Feb 10, 2017

$d = d_0 + v_it + \frac12at^2$

Plugging in all of our values, we get

$d = 1 + 0 + \frac12(4)(5^2)$

$d = 51\ m$

Denton Young
Feb 10, 2017

accel = 4.

Velocity, integrating = $4x$ + C. C is given as 0.

Position is $2x^2$ + C. C is given as 1. 2(25) + 1 = 51.

Bar Kam
Apr 2, 2018

Simple series:

4+8+12+16+20=50

-Every second the velocity changes by 4 m/s

-Every second the particle moves v units...

So, 50+1=51

1- starting position