Question

A body moves along one dimension with a constant acceleration of 3.25 m/s2 over a time interval. At the end of this interval, it has reached a velocity of 13.6 m/s.(a)If its original velocity is 6.80 m/s, what is its displacement (in m) during the time interval? m(b)What is the distance it travels (in m) during this interval? m(c)A second body moves in one dimension, also with a constant acceleration of 3.25 m/s2 , but over some different time interval. Like the first body, its velocity at the end of the interval is 13.6 m/s, but its initial velocity is −6.80 m/s. What is the displacement (in m) of the second body over this interval? m(d)What is the total distance traveled (in m) by the second body in part (c), during the interval in part (c)? m

Answer 1

a = acceleration = 3.25 m/s^2

vf = final velocity = 13.6 m/s

a)

vo= initial velocity = 6.80 m/s

Δx = (displacement)

Apply:

vf^2 = vo^2 + 2aΔx

13.6^2 = 6.80^2 + 2 * 3.25 * Δx

184.96 = 46.24 +6.5Δx

184.96 - 46.24= 6.5Δx

138.72/6.5= Δx

Δx= 21.34 m

b) Distance = 21.34m ( always positive)

c)

a= 3.25 m/s^2

Vf= 13.6m/s

Vi= -6.80 m/s

vf^2 = vo^2 + 2aΔx

(13.6)^2 = (-6.80)^2 + 2 (3.25) Δx

184.96 = 46.24 + 6.5 Δx

184.96 - 46.24 = 6.5Δx

138.72 /6.5 =

Δx= 21.34m

d)

vf^2 = vo^2 + 2Δx

0 = (-6.80) ^2 + 2 (3.25) Δx

0 = 46.25 + 6.5 Δx

-46.25 = 6.5 Δx

Δx = -46.25/6.5

Δx = -7.12m

vf^2 = 0^2 + 2aΔx

13.6^2= 2*3.25Δx

184.96 = 6.5 Δx

184,96/6.5 =Δx

Δx=28.46 m

Add both displacements ( since distance is always positive)

7.12 + 28.46= 35.58 m