If a car with velocity #100 "km"/"hr"# slams on the brakes with constant deceleration of #10 "m"/("s"^2)#, how far does the car travel before coming to a complete stop?

Using integrals as general (and particular) solutions to a first order, linear, differential equation.

2 Answers
Jun 13, 2017

#approx2.78sec# or #100/36sec#

Explanation:

Using #suvat# equations,
#v=u+at#
#0=(100*1000)/(60*60)+(-10)t#
#10t=100000/3600#
#t=100/36secapprox2.78sec#

Jun 13, 2017

#t=2.78" s"#

Explanation:

Deceleration is just negative acceleration.

Acceleration is #a(t)=-10"m"/("s"^2)#

Velocity is the antiderivative of acceleration.

#v(t)=int-10dt#

#v(t)=-10t+C_1#

Because the initial velocity at #t=0# was #100"km"/"h"#, but acceleration is in meters per second, per second, we need to convert this velocity to meters per second.

#100cancel("km")/cancel("h")xx(1000" m")/(1cancel(" km"))xx(1cancel(" h"))/(3600" s")=27.8 "m"/"s"#

So, at time #t=0#, we have

#v(0)=-10(0)+C_1=27.8#

#=> C_1=27.8#

#v(t)=-10t+27.8#

We want to know the time when the car stops, which means when the velocity is zero.

#v(t)=-10t+27.8=0#

#27.8=10t#

#t=2.78" s"#