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STALIN [3.7K]
4 years ago
15

A car traveling at 91.0 km/h approaches the turn off for a restaurant 30.0 m ahead. If the driver slams on the brakes with the a

n acceleration of -6.40m/s^2what will her stopping distance
Physics
1 answer:
eduard4 years ago
7 0

Answer: 49.92 m

Explanation:

In this situation the following equation will be useful:

V^{2}=V_{o}^{2} +2 a d

Where:

V=0 m/s is the final velocity of the car, when it finally stops

V_{o}=91 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=25.27 m/s is the initial velocity of the car

a=-6.4 m/s^{2} is the constant acceleration of the car after the driver slams on the brakes

d is the stopping distance

Isolating d:

d=\frac{-V_{o}^{2}}{2a}

d=\frac{-(25.27 m/s)^{2}}{2(-6.4 m/s^{2})}

d=41.919 m \approx 41.92 m

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A 0.5 kg basketball moving 5 m/s to the right collides with a 0.05 kg tennis
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Answer:

A. 1.4 m/s to the left

Explanation:

To solve this problem we must use the principle of conservation of momentum. Let's define the velocity signs according to the direction, if the velocity is to the right, a positive sign will be introduced into the equation, if the velocity is to the left, a negative sign will be introduced into the equation. Two moments will be analyzed in this equation. The moment before the collision and the moment after the collision. The moment before the collision is taken to the left of the equation and the moment after the collision to the right, so we have:

M_{before} = M_{after}

where:

M = momentum [kg*m/s]

M = m*v

where:

m = mass [kg]

v = velocity [m/s]

(m_{1} *v_{1} )-(m_{2} *v_{2})=(m_{1} *v_{3} )+(m_{2} *v_{4})

where:

m1 = mass of the basketball = 0.5 [kg]

v1 = velocity of the basketball before the collision = 5 [m/s]

m2 = mass of the tennis ball = 0.05 [kg]

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v3 =  velocity of the basketball after the collision [m/s]

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Now replacing and solving:

(0.5*5) - (0.05*30) = (0.5*v3) + (0.05*34)

1 - (0.05*34) = 0.5*v3

- 0.7 = 0.5*v

v = - 1.4 [m/s]

The negative sign means that the movement is towards left

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3 years ago
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F=ma
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