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.