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alina1380 [7]
3 years ago
7

With good tires and breaks, a car traveling 50 mi/hr on dry pavement can travel 400 ft when the driver reacts to something he se

es and brings the vehicle to a stop. If this is done uniformly, what is the car's acceleration?
Physics
1 answer:
Bogdan [553]3 years ago
3 0

Answer:6.72 m/s^2

Explanation:

Given

initial velocity u=50 mi/hr\approx 73.33 ft/s

Distance traveled before stopping s=400 ft

using equation of

v^2-u^2=2as

where v=final velocity

u=initial Velocity

a=acceleration

s=displacement

v=0 as car stops after travelling 400 ft

0-73.33^2=2\times a\times 400

a=-6.72 m/s^2

negative sign indicates it is deceleration

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Answer and Explanation

Arranging the measured values in increasing order;

4.3s, 4.6s, 4.6s, 4.8s, 5.1s, 5.8s

The two outliers are obviously 4.3s and 5.8s; An outlier is a value in a statistical sample which does not fit a pattern that describes most other data point. Outliers make the average value complicated. So, it is usually better for data to be precise with data points spreading out around a small area.

So, the mean is the average of the four remaining data points after removing the outliers.

Mean = (4.6 + 4.6 + 4.8 + 5.1)/4

Mean = 4.775s

So, the value recorded should be 4.775s, 4.78s or 4.8s depending on the number of decimal places allowed.

QED!

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3 years ago
In April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min, 43.6 s. Suppose "Pre" was at the 8.13 km mark at a
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Answer:0.084 m/s^2

Explanation:

Given

Total time=27 min 43.6 s=1663.6 s

total distance=10 km

Initial distance d_1=8.13 km

time taken=25 min =1500 s

initial speed v_1=\frac{8.13\times 1000}{25\times 60}=5.6 m/s

after 8.13 km mark steve started to accelerate

speed after 60 s

v_2=v_1+at

v_2=5.6+a\times 60

distance traveled in 60 sec

d_2=v_1\times 60+\frac{a60^2}{2}

d_2=336+1800 a

time taken in last part of journey

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distance traveled in this time

d_3=v_2\times t_3

d_3=\left ( 5.6+a\times 60\right )103.6

and total distance=d_1+d_2+d_3

10000=8.13\times 1000+336+1800 a+\left ( 5.6+a\times 60\right )103.6

1870=336+1800 a+\left ( 5.6+a\times 60\right )103.6

a=0.084 m/s^2

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