Suppose the position vector for a particle is given as a function of time by r(t

Question

Suppose the position vector for a particle is given as a function of time by r(t)-x(t) + y(t), with x(t-at + b and y(t,-ct2 + d, where a 1.80 m/s,1.30 m,0.124 m/s2, and d1.04 m (a) Calculate the average velocity during the time interval fromt2.15 s tot3.90 s. v-(1.80)i(2.59 xm/s (b) Determine the velocity at t2.15 s -1.80)(0.428X m/s Determine the speed at t 2.15 s 2.043 Your response is within 10% of the correct value. This may be due to roundoff error r you could have Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. m/s mistake in your calculat! n.

Explanation / Answer

here,

r = x(t) i + y(t) j

r = ( at + b) i + ( ct^2 + d) j

putting the values of constants

r = ( 1.8 t + 1.3 ) i + ( 0.124 * t^2 + 1.04) j ....(1)

at t = 2.15 s

r(2.15) = ( 1.8 * 2.15 + 1.3 ) i + ( 0.124 * 2.15^2 + 1.04) j

r(2.15) = 5.17 i + 1.61 j

at t = 3.9 s

r(3.9) = ( 1.8 * 3.9 + 1.3 ) i + ( 0.124 * 3.9^2 + 1.04) j

r(3.9) = 8.32 i + 3.93 j

a)

the average velocity during the time interval t = 2.15 s to 3.9 s , v = ( r(3.9) - r(2.15)) / (3.9 - 2.15)

v = 1.8 i m/s + 1.33 j m/s

b)

differentiating equation (1)

v(t) = dr(t)/dt

v(t) = 1.8 i + 0.248 t j

at t = 2.15 s

v(2.15 ) = 1.8 i + 0.248 * 2.15 j

v(2.15) = 1.8 i m/s + 0.53 j m/s

c)

the speed at t = 2.15 s , |v| = sqrt(1.8^2 + 0.53^2) = 1.88 m/s

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