Acar braked with a constant deceleration of 16ft/s2, producing skid marks measuring 200 feet before coming to a stop. how fast was the car traveling (in ft/s) when the brakes were first applied

the car was traveling at a speed of 80 ft/s when the brakes were first applied.

Step-by-step explanation:

The car braked with a constant deceleration of 16ft/s^2. This is a negative acceleration. Therefore,

a = - 16ft/s^2

While decelerating, the car produced skid marks measuring 200 feet before coming to a stop.

This means that it travelled a distance,

s = 200 feet

We want to determine how fast the car was traveling (in ft/s) when the brakes were first applied. This is the car's initial velocity, u.

Since the car came to a stop, its final velocity, v = 0

Applying Newton's equation of motion,

v^2 = u^2 + 2as

0 = u^2 - 2 × 16 × 200

u^2 = 6400

u = √6400

u = 80 ft/s

the system of inequity is

step-by-step explanation:

from the given graph it is noticed that we have two related lines. one is solid and another is dashed.

if a line passing through two points, then the equation of line is

the solid line passing through the points (0,4) and (4,0).

the related equation of solid line is .

similarly, find the related equation of dashed line. the dashed liune passing through the points (0,0) and (3,1).

the related equation of dashed line is

from the given graph it is noticed that the point (0,1) lies on the shaded region, therefore the point (0,1) must be satisfied by the related equations.

check the related equations by the point (0,1).

this statement is true if the sign of inequality is < instead of equality. since it is an solid line therefore the sign of inequality will be . the required inequality is

similarly, check the second related equation.

it is a dashed line, so this statement is true if the sign of inequality is > . the required inequality is