A compact car, mass 755 kg, is moving at 1.00 ✕ 102 km/h toward the east. The driver of the compact car suddenly applies the brakes hard for 1.8 s. As a result, an average force of 5.8 ✕ 103 N is exerted on the car to slow it down. What is the final velocity of the car?
To find the final velocity of the car, we can use Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration.
First, let's convert the initial velocity from km/h to m/s. We know that 1 km/h is equal to 0.2778 m/s. So, the initial velocity of the car is:
V₁ = 1.00 × 10² km/h × 0.2778 m/s/km/h
= 277.8 m/s
Next, we need to calculate the acceleration of the car. We can use the equation:
F = m × a
Where F is the force, m is the mass, and a is the acceleration.
The force exerted on the car is given as 5.8 × 10³ N, and the mass of the car is 755 kg. Plugging the values into the equation, we can solve for acceleration:
5.8 × 10³ N = 755 kg × a
a = (5.8 × 10³ N) / (755 kg)
≈ 7.6821 m/s²
Finally, we can use the equation of motion to find the final velocity of the car. The equation is:
V² - V₁² = 2aΔx
Where V is the final velocity, V₁ is the initial velocity, a is the acceleration, and Δx is the displacement.
Since the car is coming to a stop, the final velocity will be zero. We assume that the car starts braking from 0 displacement.
0 - (277.8 m/s)² = 2 × 7.6821 m/s² × Δx
Simplifying the equation:
0 - 77163.24 m²/s² = 15.3642 m/s² × Δx
Δx = (0 - 77163.24 m²/s²) / (15.3642 m/s²)
= -5023.8 m
We obtain a negative displacement since the car is moving in the east direction.
Finally, since the displacement is negative, the final velocity of the car will be in the opposite direction of the initial velocity. Therefore, the final velocity of the car is approximately -38.19 m/s in the east direction.
a=F/m
v=v(ini)-at
km/h -> m/s !!!!