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dangina [55]
3 years ago
7

An object moving on the x axis with a constant acceleration increases its x coordinate by 82.9 m in a time of 2.51 s and has a v

elocity of 20 m/s at the end of this time. Determine the acceleration of the object during this motion.
Physics
1 answer:
Aneli [31]3 years ago
5 0

We are given: Final velocity (v_f)=20 m/s .

Time t= 2.51 s and

distance s = 82.9 m.

We know, equation of motion

v_f = v_i + at.

Let us plug values of final velocity, and time in above equation.

20=v_i+a(2.51)

20=v_i+2.51a

Subtracting 2.51a from both sides, we get

20-2.51a=v_i  -----------equation(1)

Using another equation of motion

v_f-v_i=2as

Plugging values of vi =20-2.51a, t=2.51 and distnace s=82.9 in this equation.

We get,

20-(20-2.51a)=2*a(82.90)

Now, we need to solve it for a.

20-20+2.51a=165.8a.

-163.29a=0

a=0.

So, the acceleration would be 0 m/s^2.


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