A car braked with a constant deceleration of 40 ft/s2, producing skid marks measuring 160 ft before coming to?

Question

A car braked with a constant deceleration of 40 ft/s2, producing skid marks measuring 160 ft before coming to a stop. How fast was the car traveling when the brakes were first applied?

Answer 1

The braking force was enough to dissipate the kinetic energy.

Force x distance = work = kinetic energy

ma x 160 = mv^2/2 , notice that m cancels out

160a = v^2/2, solve for v...

Answer 2

There is a formula that relates the initial and final velocities of an object, given the acceleration and distance travelled. We have all of those things except for the initial velocity, which we are trying to find. Very convenient!

v(final)^2 = v(initial)^2 + 2(acceleration)(distance)

v(final) = 0, acceleration = -40, distance = 160.

0 = v(initial)^2 + 2(-40)(160)
v(initial)^2 = 12800
v(initial) = 113.14 ft/s.

This corresponds to a speed of 77.14 mph. I don't know a whole lot about cars, but I think this car had okay brakes :D

Answer 3

practice V^2- U^2 = 2aS , the position V is very last Velosity Which for sure is 0 . U is initial Velosity that is requested. a is the acceleration that is given - 40 ft/sec^2 .S is the area travelled that is given one hundred and sixty ft . therefore through putting those values weget -U^2 = 2X(-40)X160 or, U^2 = 80X160 or, U^2 = 80X80X2 therefore U= 80Xsqrt 2 ft/sec

Answer 4

Use the velocity under constant acceleration equation to solve this problem.

Let Vf = 0 ft/s
Let a = -40 ft/s
Let d = 160 ft
Vo = ?

Eqn: vf^2 = vo^2 + 2*a*d

1) 0^2 = vo^2 + 2*-40*160
2) 0 = vo^2 + -12800
3) vo^2 = 12800
4) vo = sqrt 12800
5) vo = about 113.137 ft/s

Answer 5

0=u^2-2*40*160

u=rt12800
=80rt2ft/sec