Consider two cars, a 700kg Porsche and a 600kg Honda Civic. The Porsche is speeding along at 40 m/s (mph) and the Civic is going half the speed at 20 m/s. If the two cars brake to a stop with the same constant acceleration, lets look at whether the amount of time required to come to a stop or the distance traveled prior to stopping is influenced by their initial velocity.
1. A car traveling 5m/s slams on its brakes, creating an acceleration of -2 m/s^2. How far did the car travel after it applied its brakes?
2. The same car traveling for 10m/s applies the same acceleration of -2 m/s^2. How far did the car travel after it applied its brakes?

Explanation:

To find the distance covered by the car after it applied brakes, we use 3rd equation of motion.

2as = Vf² - Vi²

s = (Vf² - Vi²)/2a

1.

We have:

Vi = Initial Velocity = 5 m/s

Vf = Final Velocity = 0 m/s    (Since, car finally stops)

a = deceleration = - 2 m/s²

s = distance covered by the car = ?

Therefore,

s = [(0 m/s)² - (5 m/s)²]/2(- 2 m/s²)

s = 6.25 m

2.

We have:

Vi = Initial Velocity = 10 m/s

Vf = Final Velocity = 0 m/s    (Since, car finally stops)

a = deceleration = - 2 m/s²

s = distance covered by the car = ?

Therefore,

s = [(0 m/s)² - (10 m/s)²]/2(- 2 m/s²)

s = 25 m

Hence, the distance traveled by the car is affected by the initial speed in accordance with a direct relationship.

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6

if the person is shown to have +90n, then ans is -10n.

if the person is shown to have -90n, then ans is +10n.