The motion of an object undergoing constant acceleration can be modeled by the kinematic equations. One such equation is xf=xi+vit+12at2 where xf is the final position, xi is the initial position, vi is the initial velocity, a is the acceleration, and t is the time. Let's say a car starts with an initial speed of 15 m/s, and moves between the 1000 m and 5000 m marks on a roadway in a time of 60 s. What is its acceleration?

Question

Echo

Physics

19 July, 23:04

The motion of an object undergoing constant acceleration can be modeled by the kinematic equations. One such equation is xf=xi+vit+12at2 where xf is the final position, xi is the initial position, vi is the initial velocity, a is the acceleration, and t is the time. Let's say a car starts with an initial speed of 15 m/s, and moves between the 1000 m and 5000 m marks on a roadway in a time of 60 s. What is its acceleration?

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Answers (1)

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Deangelo Stevenson · Answer 1

a = 1.72 m/s²

Explanation:

The given kinematic equation is the 2nd equation of motion. The equation is as follows:

xf = xi + (Vi) (t) + (1/2) (a) t²

where,

xf = the final position = 5000 m

xi = the initial position = 1000 m

Vi = the initial velocity = 15 m/s

t = the time taken = 60 s

a = acceleration = ?

Therefore,

5000 m = 1000 m + (15 m/s) (60 s) + (1/2) (a) (60 s) ²

5000 m = 1000 m + 900 m + a (1800 s²)

5000 m = 1900 m + a (1800 s²)

5000 m - 1900 m = a (1800 s²)

a (1800 s²) = 3100 m

a = 3100 m/1800 s²

a = 1.72 m/s²