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Vitek1552 [10]
3 years ago
8

You are driving down the highway late one night at 20 m/s when a deer steps onto the road 35m in front of you. Your reaction tim

e before stepping on the brakes is 0.50s, and the maximum deceleration of your car is 10m/s^2. How much distance is between you and the deer when you come to a stop?Using the information given above and the value for distance that you found above, determine the time required for you to stop once you press the brakes.
Physics
1 answer:
8090 [49]3 years ago
6 0

Answer:

- Distance between car and the deer when the car stopped = 20 m

- The time required for you to stop once you press the brakes = less than 5 s in order not to hit the deer.

Explanation:

Using the equations of motion,

In the 0.5 s reaction time, we need to first calculate how far he has travelled in that time.

a = 0 m/s² (Since the car is travelling at constant velocity)

x = ?

Initial velocity = u = 20 m/s

x = ut + at²/2

x = 20×0.5 + 0 = 10 m

From that moment,

a = - 10 m/s²

u = initial velocity at the start of the deceleration = 10 m/s

v = final velocity = 0 m/s

x = ?

v² = u² + 2ax

0² = 10² + 2(-10)(x)

20x = 100

x = 5 m

Total distance travelled from when the deer stepped onto the road = 10 + 5 = 15 m

Distance between car and the deer when the car stopped = 35 - 15 = 20 m

b) To determine the time required to stop once you step on the brakes

u = 10 m/s

t = ?

v = 0 m/s²

x = distance from when the brake was stepped on to the deer = 35 - 10 = 25 m

x = (u + v)t/2

25 = (10 + 0)t/2

10t = 50

t = 5 s

Meaning the time required to stop once you step on the brakes is less than 5s.

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Explanation:

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The problem is also an illustration of elastic collision where there is no loss in kinetic energy.

Equation (1) is a mathematical representation of the the principle of conservation of linear momentum for two colliding bodies of masses m_1 and m_2 whose respective velocities before collision are u_1 and u_2;

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m_1=1.3kg\\u_1=5m/s\\m_2=1.3kg\\u_2=0m/s

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