A car of mass 1200 kg, travelling at a steady speed of 15 m/s performs an emergency stop which takes 4.5 seconds. What braking force was applied by the driver and car to achieve this?

4000 N

Step-by-step explanation:

this question should be in physics actually

m = 1200 kg

initial velocity = vo = 15 m/s

final velocity = vt = 0 because it stops

time needed = t = 4.5

so we can find the Force F by these step

first find deacceleration

vt = vo + at

0 = 15 + a(4.5)

15 = -4.5a

a = - 15/4.5 = - 10/3 = - 3.33 m/s²

F = m.a = 1200 . 10/3 = 4000 N

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!