Question

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 velocity of 20 m/s at the end of this time. Determine the acceleration of the object during this motion.

Answer 1

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.