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Lelu [443]
3 years ago
14

A car came to a stop from a speed of 35 m/s in a time of 8.1 seconds. What was the acceleration of the car?

Physics
2 answers:
uranmaximum [27]3 years ago
5 0
Simply subtract the two velocities and divide by 8.1,

\frac{0 - 35}{8.1} = - 4.32

~~

I hope that helps you out!!

Any more questions, please feel free to ask me and I will gladly help you out!!

~Zoey
Ne4ueva [31]3 years ago
4 0

Answer:

The acceleration of the car is -4.32 m/s^2.

(C) is correct.

Explanation:

Given that,

Speed of car u= 35 m/s

Time t = 8.1 s

We need to calculate the acceleration

using equation of motion

v = u+at

Where, v = final velocity

u = initial velocity

t = time

a = acceleration

0=35+a\times8.1

a = -\dfrac{35}{8.1}

a = -4.32\ m/s^2

Hence, The acceleration of the car is -4.32 m/s^2.

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A body moving in the positive x direction passes the origin at time t = 0. Between t=0 and t=1 second, the body has a constant s
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The important thing in this type of questions is to keep track of what you are doing at every point.

From t=0 to t=1s, there is uniform motion at v=24 m/s.

Therefore, it will move 24m in that second of time.

Afterwards, an acceleration comes in, so from t=1s to 11s there will be uniformly accelerated motion with acceleration -6m/s^2.

To account for this, you need to use Suvat's equation:

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You should know what to plug in for each of the symbols in this equation:

x0 [initial position at t=1s] = 24m

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