A car weighing 9800 N travels at 30 m/s. What braking force (to the nearest whole number) brings it to rest in 100 m???? THNX BOBPURSLeY

force*distance= 1/2 mass*velocityinitial^2

solve for force.

4500

your wrong

To find the braking force required to bring the car to rest, we can use the equation:

Force (F) = Mass (m) x Acceleration (a).

First, let's find the mass of the car. We can use the equation:

Weight (W) = Mass (m) x Gravity (g).

Given that the weight of the car is 9800 N, and assuming the acceleration due to gravity is 9.8 m/s², we can solve for the mass:

m = W / g
m = 9800 N / 9.8 m/s²
m ≈ 1000 kg

Now, let's calculate the deceleration required to bring the car to rest using the equation:

Acceleration (a) = Change in velocity (Δv) / Time (t).

The change in velocity can be calculated by subtracting the final velocity (0 m/s) from the initial velocity (30 m/s):

Δv = Vf - Vi
Δv = 0 m/s - 30 m/s
Δv = -30 m/s

Since the car must come to rest in 100 m, we can find the time it takes using the equation:

Distance (d) = Initial velocity (Vi) x Time (t) + 0.5 x Acceleration (a) x Time² (t²).

Substituting the given values:

100 m = 30 m/s x t + 0.5 x a x t²

Now, we have two equations:

Δv = a x t
100 m = 30 m/s x t + 0.5 x a x t²

Rearranging the first equation:

t = Δv / a

Substituting into the second equation:

100 m = 30 m/s x (Δv / a) + 0.5 x a x (Δv / a)²
100 m = 30 m/s x Δv / a + 0.5 x Δv² / a

Now, let's substitute the value for Δv:

100 m = 30 m/s x (-30 m/s) / a + 0.5 x (-30 m/s)² / a
100 m = -900 m²/s / a + 450 m²/s² / a

To simplify, we can multiply both sides by a:

100 m x a = -900 m²/s + 450 m²/s²

Now, we can rearrange the equation to solve for acceleration:

a x 100 m = -900 m²/s + 450 m²/s²
a x 100 m = -450 m²/s + 450 m²/s²
a x 100 m = 450 m²/s² - 450 m²/s

Now, let's simplify the equation and solve for acceleration:

a x 100 m = 0
a = 0 / 100 m
a = 0 m/s²

Since acceleration is 0 m/s², the braking force required to bring the car to rest is also 0 N.