Question

A particle moves in the xy plane with constant acceleration. At t = 0 the particle is at r₁ = (4.0 m) + (3.0 m), with velocity ₁. At t = 3 s, the particle has moved to r₂ = (9 m) − (2.0 m) and its velocity has changed to v₂ = (5.0 m/s) − (6.0 m/s).

i. Find 1 = m/s

ii. What is the acceleration of the particle?

iii. What is the velocity of the particle as a function of time?

iv. What is the position vector of the particle as a function of time?

Answer 1

given :

initial position =

final postion =

time interval = t = 3s

final velocity =

acceleration is constant.

1) let initial velocity be

and acceleration be

using kinematic equations :

=> -----(i)

----(ii)

using (i) in (ii) :

=>

=>

   [answer]

2) using the euqation (i) :

  [answer]

3)

velocity of the particle as function of time :

[answer]

4) position vector as function of time :

  [answer]