The initial speed of a certain car is #"10 m/s"# before it brakes over the course of #"10 m"#. If its initial speed was #"30 m/s"# instead, what distance would it need to successfully brake with the same deceleration?
1 Answer
Intuitively, without doing much, since
Well, let's look at what variables you have.
#v_i = "10 m/s"# #v_f = "0 m/s"# #Deltax = "10 m"#
From those variables, recall the equations:
#\mathbf(v_(fx)^2 = v_(ix)^2 + 2baraDeltax)# where:
#v_(fx)# and#v_(ix)# are the final and initial velocities in the x direction, respectively.#bara# is the AVERAGE acceleration.#Deltax# is the distance in the x direction.and
#\mathbf(sumF = F_"brk" = mbara)# where:
#bara < 0# because the average acceleration is in the negative x direction (because you slow down...)#m# is the mass of the car, which we don't need to know.#F_"brk"# is the braking force.
Solving for the average acceleration, we get:
#color(green)(bara) = -(v_(ix)^2)/(2Deltax)#
#= -("10 m/s")^2/(2("10 m"))#
#= color(green)(-"5 m/s"^2)#
From this, we'd find the expression for the braking force in situation number one. Since we don't know the mass of the car, we probably don't have to know either the braking force nor the mass of the car.
Obviously the mass of the car is constant. We know that. Surprisingly enough, that is important.
Also, we are saying that
#F_"brk1" = F_"brk2" = mbara_1 = mbara_2#
Simply from that, and the fact that the car can't magically change mass, we can deduce that
Now, let's solve for
#color(blue)(Deltax) = -(v_(ix)^2)/(2bara)#
#= -("30 m/s")^2/(2(-"5 m/s"^2))#
#=# #color(blue)("90 m")#
