The driver of a car slams on the brakes when he sees a tree blocking the road. the car slows uniformly with acceleration of −5.65 m/s2 for 4.00 s, making straight skid marks 62.9 m long, all the way to the tree. with what speed (in m/s) does the car then strike the tree?

the speed with which the car strikes the tree is 27.025 m/s

Step-by-step explanation:

We would apply Newton's equation of motion. It is expressed as

S = ut + 1/2at²

Where

S represents the distance covered by the car

u represents the initial velocity of the car( the speed with which the car strikes the tree)

t represents the time taken to cover the distance

a represents the deceleration of the car.

From the information given,

S = 62.9 meters

a = - 5.65 m/s2

t = 4 seconds

Therefore

62.9 = u × 4 + 1/2 × - 5.65 × 4²

62.9 = 4u - 45.2

4u = 62.9 + 45.2

4u = 108.1

u = 108.1/4 = 27.025 m/s

were is the equation

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3