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Pavel [41]
3 years ago
8

You are traveling on an interstate highway at the posted speed limit of 70 mph when you see that the traffic in front of you has

stopped due to an accident up ahead. You step on your brakes to slow down as quickly as possible. Assume that you to slow down to 30 mph in about 5 seconds. A) With this same average acceleration, how much longer would it take you to stop?
B) What total distance would you travel from when you first apply the brakes until the car stops?
Physics
1 answer:
vovikov84 [41]3 years ago
3 0

Answer:8.75 s,

136.89 m

Explanation:

Given

Initial velocity=70 mph\approx 31.29 m/s

velocity after 5 s is 30 mph\approx 13.41 m/s

Therefore acceleration during these 5 s

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

a=\frac{13.41-31.29}{5}=-3.576 m/s^2

therefore time required to stop

v=u+at

here v=final velocity =0 m/s

initial velocity =31.29 m/s

0=31.29-3.576\times t

t=\frac{31.29}{3.576}=8.75 s

(b)total distance traveled before stoppage

v^2-u^2=2as

0^2-31.29^2=2\times (-3.576)\cdot s

s=136.89 m

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Answer: concave lens

Explanation:

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4 0
3 years ago
1.Mention two uses of the concave mirror.
Anuta_ua [19.1K]

Answer:

1. telescope

2.

f =  \frac{r}{2}

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8 0
3 years ago
What percentage of the original kinetic energy is convertible to internal energy?
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7 0
3 years ago
Raindrops fall vertically at 7.5 m/s relative to the Earth. What does an observer in a car moving at 20.2 m/s in a straight line
Vilka [71]

Answer:

vDP = 21.7454 m/s

θ = 200.3693°

Explanation:

Given

vDE = 7.5 m/s

vPE = 20.2 m/s

Required:  vDP

Assume that

vDE to be in direction of - j

vPE to be in direction of i

According to relative motion concept the velocity vDP is given by

vDP = vDE - vPE     (I)

Substitute in (I) to get that

vDP = - 7.5 j - 20.2 i

The magnitude of vDP is given by

vDP = √((- 7.5)²+(- 20.2)²) m/s =  21.7454 m/s

θ = Arctan (- 7.5/- 20.2) = 20.3693°

θ is in 3rd quadrant so add 180°

θ = 20.3693° + 180° = 200.3693°

4 0
3 years ago
A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative t
enot [183]

Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it

now in part a)

if effect of wind is in opposite direction then it travels 6 km in 20 min

so the average speed is given by the ratio of total distance and total time

v_{avg} = \frac{6000}{20*60}

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now since effect of wind is in opposite direction then we can say

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5 = 9 - v_{wind}

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Part b)

now if bird travels in the same direction of wind then we will have

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now we can find the time to go back

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Part c)

Total time of round trip when wind is present

T = t_1 + t_2

T = 20 + 7.7 = 27.7 min

now when there is no wind total time is given by

T = \frac{6000}{9} + \frac{6000}{9}

T = 22.22 min

So due to wind time will be more

4 0
4 years ago
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