Question

A particle starts from the origin at t = 0 s with a velocity of 10.0 j m/s and moves in the x-y plane with a constant acceleration of (8.0 iˆ + 2.0 jˆ ) m s¯². (a) At what time is the x- coordinate of the particle 16 m? What is the y-coordinate of the particle at that time? (b) What is the speed of the particle at the time ?

Answer 1

Given: The initial velocity of the particle at t=0 s is 10 j ^ ms −1 and its acceleration is ( 8.0 i ^ + 2.0 j ^ ) ms −2 .

a)

The acceleration vector is given as,

a= dv dt dv=adt ∫ dv = ∫ adt v= ∫ adt +u

Where, u is the initial velocity, v is the velocity vector.

By substituting the given values in the above expression, we get

∫ dv = ∫ ( 8.0 i ^ +2.0 j ^ )dt +u v=8.0t i ^ +2.0t j ^ +10 j ^ =8.0t i ^ +( 10.0+2.0t ) j ^

The position vector is given as,

r= ∫ vdt

By substituting the given values in the above expression, we get

r= ∫ [ 8.0t i ^ +( 10.0+2.0t ) j ^ ]dt =4.0 t 2 i ^ +( 10.0t+ t 2 ) j ^

The time at which the x coordinate of the particle is 16 m is given by,

4.0 t 2 =16 t 2 = 16 4 =4 t=2.0 s

At t=2.0 s, the y coordinate of the particle is,

y=10.0t+ t 2 =10.0×2+ ( 2 ) 2 =24 m

Thus, the x coordinate of the particle is 16 mat t=2.0 sand the y coordinate of that particle at this time is 24 m.

b)

The velocity of the particle at any moment is given as,

v=8.0t i ^ +( 10.0+2.0t ) j ^

The velocity of the particle at time t=2 s is,

v=8.0×2 i ^ +( 10.0+2.0×2 ) j ^ =16.0 i ^ +14 j ^   ms −1

The speed of the particle is given as,

| v |= ( 16 ) 2 + ( 14 ) 2 = 256+196 =21.26  ms −1

Thus, the speed of the particle at t=2.0 s is 21.26  ms −1 .