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Furkat [3]
3 years ago
7

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 78.5 m/s. Th

e driver of the Thunderbird realizes that she must make a pit stop, and she smoothly slows to a stop over a distance of 250 m. She spends 5.00 s in the pit and then accelerates out, reaching her previous speed of 78.5 m/s after a distance of 370 m. At this point how far has the Thunderbird fallen behind the Mercedes Benz, which has continued at a constant speed?
Physics
1 answer:
Andrews [41]3 years ago
3 0

Answer:

FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

Explanation:

First you have to consider that the Ford Thunderbird (FT) follows a rectilinear motion with varying acceleration, while Mercedez Benz (MB) has a constant velocity (no acceleration). So if you finde the time spent by FT in each section, and the distance, then you will find the distance for MB.

1) Vf² = Vi² + 2ad, where Vf: final velocity, Vi: ionitial velocity, a: acceleration and d: distance.

For the first portion  (0 m/s)² = (78.5 m/s)² + 2a(250 m) ⇒

-(78.5 m/s)² / 2(250m) = a ⇒ a = -12.3 m/s².

Now, you can find the corresponding time for this section with the following formule: Vf = Vi + at ⇒ 0 m/s = 78.5 m/s + (-12.3 m/s²) t

⇒ t= (-78.5 m/s)/ (-12.3 m/s²) ⇒ t= 6.4 seconds.

2) Then FT spent 5 seconds in the pit.

3) The the FT accelerates until reach 78.5 m/s again in a distance of 370 m.

Vf² = Vi² + 2ad ⇒ (78.5 m/s)² = (0 m/s)² + 2a(370 m)

⇒ (78.5 m/s)²/ 2(370 m) = a ⇒ a = 8.3 m/s²

Then, Vf = Vi + at ⇒ 78.5 m/s = 0 m/2 + (8.3 m/s²) t

⇒ (78.5 m/s)/(8.3 m/s²) = t ⇒ t = 9.5 seconds.

4) Summarizing, the FT moves 620 meters (250 + 370 mts) in 20.9 seconds ( 6.4 s + 5 s + 9.5 s).

5) During this time, MB moves

Velocity = distance/ time ⇒ Velocity x time = Distance

⇒ Distance = (78.5 m/s) x  (20.9 seconds) ⇒ Distance = 1640.6 meters

6) Finally, the FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

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2 years ago
A stone is thrown vertically upward with a speed of 18.0 . (a)How fast is it moving when it reaches a height of 11.0 ? (b)How lo
aliina [53]
For the first part, we are looking for Vf when dy=11.0
Upward is positive, downward is negative. 
So <span>Vf = square root [2(-9.8)(11.0) + (18.0)^2] </span>
<span>Vf = 10.4 m/s your answer is correct. 

For the part b, t is equals to the time took to reach and dy is equals to 11.0
you did, </span>11= 18t m/s-(1/2) 9.8t^2 then <span>-11 + 18t- 9.8t^2. By quadratic formula, for the way down the answer is 2.9 s while on it's way up, the answer is 0.77 s</span><span>
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2 years ago
hello friends,i need your help my home work now in physics,topic:motion.40 marks +brainliest if correct .​
fenix001 [56]

Answer:

See below

Explanation:

Vertical position is given by

df = do + vo t - 1/2 a t^2      df = final position = 0 (on the ground)

                                           do =original position = 2 m

                                            vo = original <u>VERTICAL</u> velocity = 0

                                            a = acceleration of gravity = 9.81 m/s^2

THIS BECOMES

0 = 2 + 0 * t  - 1/2 ( 9.81)t^2

  to show t =<u> .639 seconds to hit the ground </u>

During this .639 seconds it flies horizontally at 10 m/s for a distance of

      10 m/s * .639 s =<u> 6.39 m </u>

5 0
1 year ago
Starting at 1.0 m/s, a cheetah runs with a constant acceleration for 4.8 s reaching a speed of 28 m/s. What is the acceleration
LenaWriter [7]

Answer:

c.5.6m/s^2

Explanation:

Initial velocity of cheetah,u=1 m/s

Time taken by cheetah =4.8 s

Final velocity of cheetah,v=28 m/s

We have to find the acceleration of this cheetah.

We know that

Acceleration,a=\frac{v-u}{t}

Where v=Final velocity of object

u=Initial velocity of object

t=Time taken by object

Using the formula

Then, we get

Acceleration, a=\frac{28-1}{4.8}=\frac{27}{4.8} m/s^2

Acceleration=a=5.6 m/s^2

Hence, the acceleration of cheetah=5.6m/s^2

5 0
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