Read the Question Below please!

Question

Suppose the position vector for a particle is given as a function of time by r (t) = x(t)î + y(t)ĵ, with x(t) = at + b and y(t) = ct² + d, where a = 2.00 m/s, b = 1.50 m, c = 0.129 m/s², and d = 1.20 m.

(a) Calculate the average velocity during the time interval from t = 1.90 s to t = 4.25 s.

:v =

_____m/s

b) Determine the velocity at t = 1.90 s.

v

= _________m/s

Determine the speed at t = 1.90 s.

 :_______m/s

Answer 1

The average velocity is the displacement divided by the time difference.

R(t) at 1.9 s = 5.3î + 1.67ĵ

R(t) at 4.5 s = 10î+ 3.53j

R(4.5) - R(1.9) = 4.7î+ 1.86ĵ / (4.25 - 1.9)s = 2î + 0.791ĵm/s . The magnitude is 2.15 m/s

Velocity at 1.9 s = d R(t)/dt = 2î + 0.258 x 1.9 ĵ = 2î + 0.490ĵ

Speed is magnitude of 2î+ 0.490 ĵ = 2.06 m/s

Answer 2

Suppose the position vector for a particle is given as a function of time by r (t) = x(t)î + y(t)ĵ, with x(t) = at + b and y(t) = ct2 + d, where a = 2.00 m/s, b = 1.50 m, c = 0.129 m/s2, and d = 1.20 m.

(a) Calculate the average velocity during the time interval from t = 1.90 s to t = 4.25 s.

:v = 2.15_____m/s

b) Determine the velocity at t = 1.90 s.

v = _2.059________m/s

r(t) = (at + b)i + ( ct^2 + d ) j = ( 2t + 1.5)i + ( 0.129t^2 + 1.20)j



ave speed = [r(4.25) - r(1.90)]/ (4.25 - 1.90 )



r( 4.25 ) = ( 2*4.25 + 1.5)i + ( 0.129 * 4.25^2 + 1.20 ) = 10i + 3.53j



r( 1.90 ) = ( 2*1.90 + 1.5 )i + ( 0.129 * 1.90^2 + 1.20) = 5.3i + 1.67j



ave speed = [r(4.25) - r(1.90)]/ (4.25 - 1.90 ) = [ 10i + 3.53j - 5.3i -1.67j] / ( 2.35) ------->

 [4.7i + 1.86j]/ 2.35 = 2i + 0.79j



|v| = √( 2^2 + 0.79^2 ) = 2.15

--------------------------------------------------------------------------------------------------------

v(t) = lim ( Δ r ) / d t ----------> [( 2t + 1.5)i + ( 0.129t^2 + 1.20)j] / dt ------> v'(t) = 2i + 0.258tj

 v( 1.90) = 2i + 0.258(1.90)j

= 2i + 0.49j



|v| = √( 2^2 + 0.49^2)

 = 2.059

I hope it is useful,

Minoo√