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Natasha_Volkova [10]
3 years ago
13

A driver is traveling along a straight road at the speed limit of 60 mph. After two minutes, she slows at a constant rate to a s

top at a stop light. Two minutes later, the light turns green and she accelerates at a constant rate back up to the speed limit. Three and a half minutes later, she again slows at a constant rate to a stop. After three minutes, she performs a U-turn, then accelerates at a constant rate back up to 60 mph. Two minutes later, she reaches her destination and slows at a constant rate to a stop. Assume that each period of slowing down and speeding up lasts 30 s and that the driver is initially moving in the +x- direction. Create a graph of the driver's velocity versus time that represents her trip.

Physics
1 answer:
Yuri [45]3 years ago
7 0

Explanation :

From the given information, the graph is plotted.

It is given that, a driver is traveling along a straight road at the speed limit of 60 mph. Initially, he was at point A.

After two minutes, she slows at a constant rate to a stop at a stop light. BC shows this part.

Two minutes later, the light turns green and she accelerates at a constant rate back up to the speed limit. CD shows this part.

Three and a half minutes later, she again slows at a constant rate to a stop. DE shows this part.

Hence, this is the required solution.

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9. A Veggie meatball with v = 5.0 m/s rolls off a 1.0 m high table. How long does it take to hit the floor if no one sneezes? Wh
Paha777 [63]

By Considering the vertical distance and both vertical and horizontal final velocity, the time t = 0.45 s and Velocity V = 6.7 m/s

Given that a Veggie meatball with v = 5.0 m/s rolls off a 1.0 m high table.

Height h = 1.0 m

As the ball rolls off the table, it will be fallen under gravity. Where

g = 9.8 m/s^{2}

Initial vertical velocity u_{y} = 0

Initial horizontal velocity u_{x} = 5 m/s

Considering the vertical distance, the formula to use to calculate the time will be;

h = ut + 1/2gt^{2}

1 = 0 + 1/2 x 9.8t^{2}

1 = 4.9t^{2}

t^{2} = 1/4.9

t = \sqrt{0.204}

t = 0.45 seconds

It takes 0.45 seconds to hit the floor if no one sneezes.

To calculate its velocity when it hits the floor, we will need to calculate for both vertical and horizontal final velocity and find the resultant velocity of the two.

Vertical component

V_{y} = U_{y} + gt

V_{y} = 0 + 9.8(0.45)

V_{y} = 4.41 m/s

Horizontal component

V_{x} = u_{x} + at

but a = 0

V_{x} = 5 m/s

Final velocity V = \sqrt{5^{2} + 4.41^{2}  }

V = 6.67 m/s

Therefore, it will hit the floor at a velocity of 6.7 m/s

Learn more here: brainly.com/question/5063616

8 0
2 years ago
The Kentucky Derby is held at the Churchill Downs track in Louisville, Kentucky. The track is one and one-quarter miles in lengt
Lera25 [3.4K]

Answer:

2.815 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

s=ut+\frac{1}{2}at^2\\\Rightarrow 1609.34=16.58\times t+\frac{1}{2}\times 0.0105\times t^2\\\Rightarrow \\\Rightarrow 4.905t^2+16.58t-1609.34=0

t=\frac{-16.58+\sqrt{308.69254}}{0.0105},\:t=\frac{-16.58-\sqrt{308.69254}}{0.0105}\\\Rightarrow t=94.25, -3252\ s

Time taken with the acceleration is 94.25 seconds

Time = Distance / Speed

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Difference in time = 97.065-94.25 = 2.815 seconds

6 0
3 years ago
A roller coaster has a "hump" and a "loop" for riders to enjoy (see picture). The top of the hump has a radius of curvature of 1
nikklg [1K]

Answer:

Part a)

F_n = 306 N

Part B)

v = 12.1 m/s

Explanation:

Part A)

At the top of the hump the force on the rider is

1) Normal force

2) weight

so here we know that

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F_n = mg - \frac{mv^2}{R}

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Part B)

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mg = \frac{mv^2}{R}

v = \sqrt{Rg}

v = \sqrt{15\times 9.81}

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