Respuesta :
Explanation:
Given that,
The initial velocity of the object, u = -11.2 m/s (due west)
Acceleration of the object, [tex]a=+2.78\ m/s^2[/tex] (due east)
Time, t = 2.03 seconds
We need to find the distance traveled by the object during this time. Let the distance covered is d. Using second equation of motion to find it as :
[tex]d=ut+\dfrac{1}{2}at^2\\\\d=-11.2\times 2.03+\dfrac{1}{2}\times (2.78)\times (2.03)^2 \\\\d=-17.007\ m[/tex]
So, the object cover a distance of 17.007 meters in the westward direction.
The object travel the distance of 17.007 meters in the westward direction.
Calculation of the distance:
Since
The initial velocity of the object, u = -11.2 m/s (due west)
Acceleration of the object, (due east) a= +[tex]2.78 m/s^2[/tex](due east)
Time, t = 2.03 seconds
So here the distance is
[tex]d = ut + 1/2at^2\\\\= -11.2 * 2.03 + 1/2 * (2.78) * (2.03)^2[/tex]
= -17.007m
Hence, The object travel the distance of 17.007 meters in the westward direction.
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