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Marizza181 [45]
3 years ago
9

While passing a slower car on the highway, i accelerate uniformly from 12 m/s to 24 m/s in a time of 10.0s. How far do you trave

l during this time and what is the magnitude of your acceleration
Mathematics
1 answer:
Taya2010 [7]3 years ago
6 0
You say that there is uniform acceleration so:

vf-vi=at  (final velocity minus initial velocity is equal to acceleration times time)

We know vf, vi, and t so we can solve for acceleration:

24-12=a10

12=10a

a=1.2

That is the acceleration, we will need to integrate with respect to time twice...

v=⌠a dt

v=at+vi  , we know a=1.2m/s^2 and vi=12m/s

v=1.2t+12, 

x=⌠1.2t+12 dt

x=1.2t^2/2+12t+xo, we can just let xo=0 for this problem...

x(t)=0.6t^2+12t

Now we know that this acceleration lasts for 10 seconds so the distance traveled in that time is:

x(10)=0.6(10^2)+12(10)

x(10)=60+120

x(10)=180 meters
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<h3>Given :-</h3>
  • Pranav ran 20.3 km more than Branda

  • Pravin ran 38.6 km

\\  \\

To find :

  • No. of kilometers did Brenda run.

\\  \\

<h3>Let :</h3>

  • No. of kilometers ran by Brenda be x.

\\  \\

<h3>Solution:</h3>

\\  \\

Equation formed:-

\\  \\

Total distance covered by Pravan = More distance covered by Pravan + distance covered by Brenda.

Therefore:-

\\  \\

\leadsto \sf38.6 = 20.3 + x

Write the equation

\\  \\

\leadsto \sf38.6 - 20.3 =x

When we transfer 20.3 to left side the positive sign (+) will change into negative sign (–)

\\  \\

\leadsto \sf x = 38.6 - 20.3

Arrange the equation because x is always represented at left side.

\\  \\

\leadsto \boxed {\pmb{\sf x = 18.3}}\star

After subtracting 38.6 with 20.3 we will get result as 18.3 .

\\  \\

\therefore \red{ \underline{  \pmb{\frak{Distance  ~ covered ~by ~Brenda~ is ~equal ~to~18.3~kilometers}}}}

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