A certain car is capable of accelerating at a uniform rate of #0.85# #m##/##s^2#. What is the magnitude of the car's displacement as it accelerates uniformly from a speed of #83# #km##/##h# to one of #94# #km##/##h#?

1 Answer
Jun 20, 2017

#Deltax = 88# #"m"#

Explanation:

We're asked to find the distance a car travels with a given constant acceleration and initial and final velocities.

To find this distance, we'll use the equation

#(v_x)^2 = (v_(0x))^2 + 2a_x(x - x_0)#

Our known quantities are

  • #a_x = 0.85"m"/("s"^2)#

  • #v_x# is the final velocity, #94"km"/"h"#, which me must convert to units of #"m"/"s"# to be consistent:

#((94cancel("km"))/(1cancel("h")))((10^3"m")/(1cancel("km")))((1cancel("h"))/(3600"s")) = color(red)(26.1"m"/"s"#

  • The initial velocity #v_(0x)#, #83"km"/"h"#, which we must also convert:

#((83cancel("km"))/(1cancel("h")))((10^3"m")/(1cancel("km")))((1cancel("h"))/(3600"s")) = color(blue)(23.1"m"/"s"#

Plugging in the known values, we have

#(color(red)(26.1"m"/"s"))^2 = (color(blue)(23.1"m"/"s"))^2 + 2(0.85"m"/("s"^2))(x - x_0)#

#682("m"^2)/("s"^2) = 532("m"^2)/("s"^2) + (1.7"m"/("s"^2))(x-x_0)#

#150("m"^2)/("s"^2) = (1.7"m"/("s"^2))(x-x_0)#

#x-x_0 = color(green)(88# #color(green)("m"#

rounded to #2# significant figures, the amount given in the problem.

During this change in velocity, the car will thus travel a distance of #color(green)(88# #sfcolor(green)("meters"#.