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"