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finlep [7]
2 years ago
5

A woman was recently given the opportunity to ride in a porsche race car on their test in Hapeville, Georgia. Below is a graph o

f just a short segment of her ride. During which segments(s) of her trip was the car stopped?

Physics
1 answer:
Zinaida [17]2 years ago
6 0

Answer:

During the segments B - C and D - E, the car stopped since the y axis is the distance and the distance stayed the same in between those segments.

For a simpler answer, the flat horizontal lines on the graph are the times when the car was stopped.

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a bus is moving with the velociity of 36 km/hr . after seeing a boy at 20 m ahead on the road, the driver applies the brake and
Klio2033 [76]

Answer:

Assumption: the acceleration of this bus is constant while the brake was applied.

Acceleration of this bus: approximately \left(-6.0\; \rm m \cdot s^{-2}\right).

It took the bus approximately 1.7\;\rm s to come to a stop.

Explanation:

Quantities:

  • Displacement of the bus: x = 10\; \rm m.
  • Initial velocity of the bus: \displaystyle u = 36\; \rm km \cdot hr^{-1} = 36\; \rm km \cdot hr^{-1}\times \frac{1\; \rm m \cdot s^{-1}}{3.6\; \rm km\cdot hr^{-1}} = 10\; \rm m \cdot s^{-1}.
  • Final velocity of the bus: v = 0\; \rm m\cdot s^{-1} because the bus has come to a stop.
  • Acceleration, a: unknown, but assumed to be a constant.
  • Time taken, t: unknown.

Consider the following SUVAT equation:

\displaystyle x = \frac{1}{2}\, \left(a\, t^2\right) + u\, t.

On the other hand, assume that the acceleration of this bus is indeed constant. Given the initial and final velocity, the time it took for the bus to stop would be inversely proportional to the acceleration of this bus. That is:

\displaystyle t = \frac{v - u}{a}.

Therefore, replace the quantity t with the expression \displaystyle \left(\frac{v - u}{a}\right) in that SUVAT equation:

\displaystyle x = \frac{1}{2}\, \left(a\, \left(\frac{v -u}{a}\right)^2\right) + u\, \left(\frac{v - u}{a}\right).

Simplify this equation:

\begin{aligned}x &= \frac{1}{2}\, \left(a\, {\left(\frac{v -u}{a}\right)}^2\right) + u\, \left(\frac{v - u}{a}\right) \\ &= \frac{1}{2}\left(\frac{{(v - u)}^2}{a}\right) + \frac{u\, (v - u)}{a} =\frac{1}{a}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right)\end{aligned}.

Therefore, \displaystyle a= \frac{1}{x}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right).

In this question, the value of x, u, and v are already known:

  • x = 10\; \rm m.
  • \displaystyle u =10\; \rm m \cdot s^{-1}.
  • v = 0\; \rm m\cdot s^{-1}.

Substitute these quantities into this equation to find the value of a:

\begin{aligned} a &= \frac{1}{x}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right) \\ &= \frac{1}{10\; \rm m}\times \left(\frac{{\left(0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}\right)}^2}{2} + \left(0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}\right)\times 10\; \rm m \cdot s^{-1}\right)\\ &\approx -6.0\; \rm m \cdot s^{-2}\end{aligned}.

(The value of acceleration a is less than zero because the velocity of the bus was getting smaller.)

Substitute a \approx -6.0\; \rm m \cdot s^{-2} (alongside u = 10\; \rm m \cdot s^{-1} and v = 0\; \rm m \cdot s^{-1}) to estimate the time required for the bus to come to a stop:

\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}}{-6.0\; \rm m \cdot s^{-2}} \approx 1.7\; \rm s\end{aligned}.

8 0
3 years ago
A 80 kg parent and a 20 kg child meet at the center of an ice rink. They place their hands together and push. The parent pushes
Ymorist [56]

Answer:

force acting on the parent = 25 N .

Explanation:

According to third law of Newton , there is equal and opposite reaction to every action . Here force by the parent on child is action and the force by child on parent is reaction . The former is given as 25 N so force by child on parent will also be 25 N .

Answer is 25 N .

4 0
2 years ago
a spring is used to launch a ball vertically into the air. the spring has a spring constant of 200N/m and is compressed by 5 cm.
zhannawk [14.2K]

Answer:

2.55 m

Explanation:

Elastic energy = gravitational energy

½ kx² = mgh

h = kx² / (2mg)

h = (200 N/m) (0.05 m)² / (2 × 0.010 kg × 9.8 m/s²)

h = 2.55 m

8 0
3 years ago
The slope of the line tangent to the curve on a position-time graph at a specific time is the
Rudik [331]

Answer:

I do I make a brinliest can you please can me

7 0
3 years ago
Convert 5.7 miles to km
gayaneshka [121]

5.7 kilometers is equal to 3.5418157957528034 miles

4 0
3 years ago
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