A car moves at a speed of 40km/h, It is stopped by applying brakes which produces a uniformacceleration of -0.6m/s2. How much distance will the vehicle move before coming to stop?
A car moves at a speed of 40 km/h. It is stopped by applying brake which produces a uniform acceleration of -0.5 m/ s². How much distance will the vehicle move before coming to stop?
V^2 = Vo^2 + 2a*d.
Vo = 40km/h = 40,000m/3600s = 11.11 m/s.
V = 0, a = -0.6m/s^2, d = ?.
d= -vo^2/2a
To find the distance the vehicle will move before coming to a stop, we first need to convert the speed from kilometers per hour (km/h) to meters per second (m/s).
Given:
Speed of the car = 40 km/h
Acceleration = -0.6 m/s²
Step 1: Convert speed from km/h to m/s
1 km = 1000 m
1 hour = 3600 seconds
Speed in m/s = (40 km/h) * (1000 m / 1 km) * (1 h / 3600 s)
Speed in m/s = (40 * 1000) / 3600
Speed in m/s ≈ 11.11 m/s
Step 2: Determine the time taken to stop
To use the equation of motion s = ut + (1/2)at², where s is the distance, u is the initial velocity, a is the acceleration, and t is the time taken, we need to find t.
Using the equation v = u + at, where v is the final velocity (0 m/s in this case), u is the initial velocity (11.11 m/s), a is the acceleration (-0.6 m/s²), and t is the time taken, we can solve for t.
0 = 11.11 m/s + (-0.6 m/s²) * t
0 = 11.11 m/s - 0.6 m/s² * t
Solving for t, we get:
0.6 m/s² * t = 11.11 m/s
t = 11.11 m/s / 0.6 m/s²
t ≈ 18.52 s
Step 3: Calculate the distance using the equation s = ut + (1/2)at²
s = (11.11 m/s) * (18.52 s) + (1/2) * (-0.6 m/s²) * (18.52 s)²
s = 205.55 m - 0.5 * 0.6 m/s² * 342.39 s²
s ≈ 205.55 m - 103.32 m
s ≈ 102.23 m
Therefore, the vehicle will move approximately 102.23 meters before coming to a stop.