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krok68 [10]
2 years ago
15

A car traveling at 40 ft/sec decelerates at a constant 5 feet per second per second. how many feet does the car travel before co

ming to a complete stop?
Physics
1 answer:
Tema [17]2 years ago
3 0
<span>This problem is solved by the equation of motion: x = x0 + v0*t + 1/2*a*t^2, Here x0 = 0, v0 = 40ft/sec and a = -5 ft/s^2, we need to solve for t: v = v0 + a*t, solve how long does it take to stop: 0 = v0 + a*t --> a*t = -v0 --> t = -v0/a -- > 40/5 = 8 seconds to stop. In this time, the car travels x = 0 + 40*8 + 0.5*-5*8^2 ft ~ 160 ft. Answer: The car travels 160 ft.</span>
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3 years ago
) a wrench 0.6 meters long lies along the positive y-axis, and grips a bolt at the origin. a force is applied in the direction o
Lera25 [3.4K]
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6 0
3 years ago
Car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver not
musickatia [10]

Answer:

Part A: t = v_0/a_0

Part B: t = v_0/a_0

Part C: v_0^2/a_0

Explanation:

Part A:

We will use the following kinematics equation:

v = v_0 + at\\0 = v_0 - a_0t\\t = \frac{v_0}{a_0}

Part B:

We will use the same kinematics equation:

v = v_0 + at\\v_0 = 0 + a_0t\\t = \frac{v_0}{a_0}

Part C:

The total time takes is 2t.

So the train moves a distance of

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And the car moves a distance in Part A and in Part B:

d_A = v_0t + \frac{1}{2}at^2 = v_0(\frac{v_0}{a_0}) - \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{a_0} - \frac{v_0^2}{2a_0} = \frac{v_0^2}{2a_0}\\d_B = v_0t + \frac{1}{2}at^2 = 0 + \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{2a_0}

So the total distance that the car traveled is d = \frac{v_0^2}{a_0}

The difference between the train and the car is

x - d = \frac{2v_0^2}{a_0} - \frac{v_0^2}{a_0} = \frac{v_0^2}{a_0}

8 0
3 years ago
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Harman [31]
Accelration=changespeed/changetime
changetime=8
changespeed=45-5=40
40/8=5

5m/s^2 is acceleration
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