The position of an object is given by x = at3 - bt2 + ct, where a = 4.1 m/s3, b = 2.2 m/s2, c = 1.7 m/s, and x and t are in SI units. What is the instantaneous acceleration of the object when t = 4.1 s?

Question

Ramiro Olsen

Physics

11 April, 23:41

The position of an object is given by x = at3 - bt2 + ct, where a = 4.1 m/s3, b = 2.2 m/s2, c = 1.7 m/s, and x and t are in SI units. What is the instantaneous acceleration of the object when t = 4.1 s?

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Answers (1)

Know the Answer?

Jaylan Wade · Answer 1

The answer to your question is: 15 m/s2

Explanation:

Equation x = at3 - bt2 + ct

a = 4.1 m/s3

b = 2.2 m/s2

c = 1.7 m/s

First we find x at t = 4.1 s

x = 4.1 (4.1) 3 - 2.2 (4.1) 2 + 1.7 (4.1)

x = 4.1 (68.921) - 2.2 (16.81) + 6.97

x = 282.58 - 36.98 + 6.98

x = 252.58 m

Now we find speed

v = x/t = 252.58 / 4.1 = 61.6 m/s

Finally

acceleration = v/t = 61.6/4.1 = 15 m/s2