John dashes from the science building with an initial velocity vector described

Question

John dashes from the science building with an initial velocity vector described by 3.50 m/s x hat + 8.50 m/s y hat. But 2.18 s later, a constant acceleration changes his velocity vector to -8.70 m/s x hat - 10.5 m/s y hat. What are the magnitude and direction of his average acceleration? John dashes from the science building with an initial velocity vector described by 3.50 m/s x hat + 8.50 m/s y hat. But 2.18 s later, a constant acceleration changes his velocity vector to -8.70 m/s x hat - 10.5 m/s y hat. What are the magnitude and direction of his average acceleration? John dashes from the science building with an initial velocity vector described by 3.50 m/s x hat + 8.50 m/s y hat. But 2.18 s later, a constant acceleration changes his velocity vector to -8.70 m/s x hat - 10.5 m/s y hat. What are the magnitude and direction of his average acceleration?

Explanation / Answer

here,

the initial speed , u = 3.5 m/s i + 8.5 m/s j

final speed , v = - 8.7 m/s - 10.5m/s j

the average accelration , a = final speed - initial speed / time taken

a = (- 8.7 m/s - 10.5m/s j - (3.5 m/s i + 8.5 m/s j)) /2.18

a = -5.6 m/s^2 - 8.72 m/s^2

|a| = sqrt(5.6^2+ 8.72^2)

|a| = 10.36 m/s^2

theta = arctan(8.72/5.6)

theta = 57.29 degree

the magnitude is 10.36 m/s^2 and in the direction 237.29 degree counterclockwise from the +x axis

Related Questions

Navigate

• Browse (All)
• Browse J
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.