A car is being driven along a country road on a dark and rainy night at a speed of 20 m/s. The section of road is horizontal and straight. The driver sees that a tree has fallen and covered the road ahead. Panicking, the driver locks the breaks at a distance of 20m from the tree. If the coefficient of friction between the wheels and road is 0.8, at what speed does the car crash into the tree?
(Friction force)*(braking distance) = K.E. reduction
Solve for the final kinetic energy and get the final V from that.
post it.
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To determine the speed at which the car crashes into the tree, we need to use the principles of Newtonian physics and specifically the concept of stopping distance.
The stopping distance of a vehicle can be calculated using the equation:
Stopping distance = Reaction distance + Braking distance
The reaction distance is the distance the car travels during the time between the driver seeing the obstacle and applying the brakes. The braking distance is the distance the car travels after the brakes are applied until it comes to a complete stop.
1. Let's calculate the reaction distance first. The driver has already seen the tree and locked the brakes at a distance of 20 meters from the tree. Therefore, the reaction distance is 20 meters.
2. Now, we need to calculate the braking distance. The braking distance can be determined using the formula:
Braking distance = (Initial velocity^2) / (2 * friction * acceleration due to gravity)
Where:
- Initial velocity is the speed of the car before the brakes are applied.
- Friction is the coefficient of friction between the wheels and road.
- Acceleration due to gravity is approximately 9.8 m/s^2.
In this case, the initial velocity is 20 m/s, and the coefficient of friction is 0.8. Plugging these values into the formula, we get:
Braking distance = (20^2) / (2 * 0.8 * 9.8)
Calculating this further:
Braking distance = 400 / (2 * 7.84)
Braking distance ≈ 25.51 meters
3. Now, we can find the stopping distance by summing up the reaction distance and the braking distance:
Stopping distance = Reaction distance + Braking distance
Stopping distance = 20 + 25.51
Stopping distance ≈ 45.51 meters
Therefore, the car will crash into the tree at a speed of 20 m/s when it reaches a distance of approximately 45.51 meters from the tree.