Physics problem please help?

Question

Acceleration- A woman driving at a speed of 23 m/s sees a deer in the road ahead and applies the brakes when she is 210 m from the deer. if the deer does not move and the car stops right before it hits the deer what is the acceleration provided by the car's brakes?

Please answer really bad at physics, show work i would like to know how to do these problems, a website would help too. Thanks

Answer 1

Use the conservation of energy. The car's initial kinetic energy is ½mv², and it's final kinetic energy is 0. The energy to stop the car is m*a*d (mass * deceleration * distance).

m*a*d = ½mv²

(m means mass, which cancel out)

a = v²/2d

a = (23 m/s)² / 2 (210 m)

a = 529 / 420

a = 1.256 m/s²

Note: The effective coefficient of friction is already factored into the deceleration.

Answer 2

a = Δv/Δt

a*t= v

210a = 23

210/23= 9.13

it depends if you want it in like kilometers per hour you need to change the 23 m/s to  */km.

Answer 3

Initial velocity u = 23 m/s

Final velocity v = 0

Displacement s = 210 m

Acceleration a = ?

v^2 = u^2 + 2as

Or, a = (v^2 - u^2)/(2s)

= (0 - 23^2)/(2 * 210)

= - 529/420

= -1.26 m/s^2

Ans: -1.26 m/s^2

anon, after writing a*t= v,

you have written

210a = 23

In other words, you have assumed t = 210. That is wrong because 210 is distance and not time.

Answer 4

Using one of the equations of motion

v^2 = u^2 + 2as, where v = final velocity

u = initial velocity

s = distance or displacement

a = acceleration

substituting the values that you gave, v = 0, because the vehicle stops

u = 23m/s

s = 210 m

0 = 23^2 + 2*a*210

rearranging the equation gives:

420 * a = ( - )23^2

a = ( - ) 529/420

a = (-)1.26 m/s^2

the minus means that the car decelerates, which must be true

therefore the acceleration is (negative) 1.26 m/s^2