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Svetradugi [14.3K]
3 years ago
8

Un auto que acelera a razón de 2 m/S^2 ; Si para cierto tramo se observa que logra triplicar su rapidez en 6 s. ¿Cuál fue su rap

idez al final de estos 6 segundos?
Physics
1 answer:
katen-ka-za [31]3 years ago
8 0

Answer:

12m/s

Explanation:

Using the equation of  motion

v = u + at

Given that

u = 0m/s

a = 2m/s²

time t = 6s

Substitute the given values into the expression

v = 0 + 2(6)

v = 0+12

v = 12m/s

Hence the required velocity is 12m/s

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4 years ago
In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th
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<u>Answer:</u>

<em>Thunderbird is 995.157 meters behind the Mercedes</em>

<u>Explanation:</u>

It is given that all the cars were moving at a speed of 71 m/s when the driver of Thunderbird  decided to take a pit stop and slows down for 250 m. She spent 5 seconds  in the pit stop.

Here final velocity v=0 \ m/s

initial velocity u= 71 m/s  distance  

Distance covered in the slowing down phase = 250 m

v^2-u^2=2as

a= \frac {(v^2-u^2)}{2s}

a = \frac {(0^2-71^2)}{(2 \times 250)}=-10.082 \ m/s^2

v=u+at

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

= \frac {(0-71)}{(-10.082)}=7.042 s

t_1=7.042 s

The car is in the pit stop for 5s t_2=5 s

After restart it accelerates for 350 m to reach the earlier velocity 71 m/s

a= \frac {(v^2-u^2)}{(2\times s)} = \frac{(71^2-0^2)}{(2 \times 370)} =6.81 \ m/s^2

v=u+at

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

t= \frac{(71-0)}{6.81}= 10.425 s

t_3=10.425 s

total time= t_1 +t_2+t_3=7.042+5+10.425=22.467 s

Distance covered by the Mercedes Benz during this time is given by s=vt=71 \times 22.467= 1595.157 m

Distance covered by the Thunderbird during this time=250+350=600 m

Difference between distance covered by the Mercedes  and Thunderbird

= 1595.157-600=995.157 m

Thus the Mercedes is 995.157 m ahead of the Thunderbird.

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Both hits the ground <u>at the same time</u> because they have <u>same vertical acceleration</u>

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<h3>What is vertical  acceleration?</h3>

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1 year ago
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