a train is moving at 45km/h when a coupling breaks and the last car separates from the train. as soon as the car separates, the brakes are automatically applied, locking all wheels of the runaway car and causing a uniform decelaration of 5 m/s^2. determine the distance the car travels before coming to a stop?
Vo = 45km/h = 45000m / 3600s=12.5 m/s = Initial velocity.
d = (Vf^2-Vo^2)/2a,
d = (0-(12.5)^2) / -10 = 15.63 m.
To determine the distance the car travels before coming to a stop, we can use the equations of motion. We'll need to use the equation of motion that relates the initial velocity (u), final velocity (v), acceleration (a), and distance (s):
v^2 = u^2 + 2as
Here's how we can apply this equation to solve the problem:
1. Convert the initial velocity from km/h to m/s:
First, we convert 45 km/h to m/s by dividing it by 3.6:
45 km/h ÷ 3.6 = 12.5 m/s (approx.)
2. Determine the deceleration:
The deceleration is given as 5 m/s^2.
3. Rearrange the equation and plug in the known values:
v^2 = u^2 + 2as
0^2 = (12.5 m/s)^2 + 2(5 m/s^2) * s
4. Simplify the equation:
0 = 156.25 + 10s
5. Solve for s:
10s = -156.25
s = -15.625 m^2/s^2
We have obtained a negative value for s, indicating that the equations of motion do not give us a positive real solution for s. This implies that the runaway car does not come to a complete stop within the given conditions.
To verify this conclusion, we should consider real-world factors that could affect the calculation, such as friction, air resistance, and the precise behavior of the brakes. It's possible that these factors may not completely stop the car within the given distance.