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4 years ago

12

A car with a mass of 1,000 kg moving with an initial velocity of 8 m/s is brought to rest by colliding with a truck. The collisi

ons takes place over a time span of 0.2 s.Calculate the force exerted on the car

1 answer:

ruslelena [56]4 years ago

4 0

Answer:

-40,000 N

Explanation:

First, use the kinematics equation v(f) = v(i) + at. Final velocity is 0, initial is 8, and time is 0.2 seconds. Solving for a, you get -40 m/s^2. Then, use Newton’s second law, F=ma, to find the force. F = (1000)(-40) = -40,000 N.

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Answers:

a) $\hat F=(0.83,-0.55) N$

b) $\hat D=(-0.44,-0.89) m$

c) $\hat V=(-0.47,0.88) m/s$

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A unit vector is a vector whose magnitude (length) is equal to 1. This kind of vector is identified as $\hat v$ and the way to calculate is as follows:

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Where:

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Having this information clarified, let's begin with the answers:

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Magnitude of $\vec F$ :

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<u />

<u>Unit vector:</u>

$\hat F=\frac{\vec F}{|F|}$

$\hat F=\frac{(9.0 \hat i - 6.0 \hat j) N}{10.81 N}$

$\hat F=\frac{9.0}{10.81} N-\frac{6.0}{10.81}N$

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Magnitude of $\vec D$ :

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<u />

<u>Unit vector:</u>

$\hat D=\frac{\vec D}{|D|}$

$\hat D=\frac{(-4.0 \hat i - 8.0 \hat j) m}{8.94 m}$

$\hat D=\frac{-4.0}{8.94} Nm+\frac{-8.0}{8.94}m$

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c) Velocity Vector

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Magnitude of $\vec V$ :

$|V|=\sqrt{(-3.50 \hat i)^{2}+(6.50 \hat j)^{2}}m/s=7.38 m/s$

<u />

<u>Unit vector:</u>

$\hat V=\frac{\vec V}{|V|}$

$\hat V=\frac{(-3.50 \hat i +6.50 \hat j) m/s}{7.38 m/s}$

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Since Power is equal to energy/time

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