A particle moves from x-y point (1,1) to point (2,4) to point (3,9) in time 2/sq

Question

A particle moves from x-y point (1,1) to point (2,4) to point (3,9) in time 2/sqr(5) seconds (.8944 seconds). It is not at rest at point (1,1).

• In what direction is the net force on it?

•What is its average acceleration in the x-direction? •

What is its average acceleration in the y-direction? (Use your basic definitions.)

• Where was it at t=0, assuming all accelerations are constant?

What was itsvelocity at t=0?

• What is its

• What is its

• Write down an expression for

r(t) using i-hat and j-ha

Explanation / Answer

a) It is following y=x^2 curve.....thus force is in +ve x direction and +ve y-direction..........

b)s= 1/2*a*t^2 implies.... (3-1) = 1/2 *a*(.8944)^2 implies avg. acceleration = 5m/s2..........

c) similarly, avg. acceleration in y dirn. = 20m/s2......

d) At (1,1).....

Displacement = sqrt(8^2+ 2^2) = 8.246m.........

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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