The graph above shows the velocity of an object vs time with the velocity given

Question

The graph above shows the velocity of an object vs time with the velocity given by:
   v = +2 m/s3t2 12 m/s2t + 0 m/s

(A) What is the (instantaneous) velocity of the object at t= 6 s?
v(6 s)= m/s [± 1.4 m/s ]
(The value in square brackets is how precise your answer needs to be to receive credit.)

(B) What is the velocity of the object at t= 5.75 s?
v( 5.75 s)= m/s. [± 1.4 m/s ]

(C) What is the average acceleration between t= 0 s and t= 8 s?
aavg= m/s2 [± 1.4 m/s2 ]
C
(D) What is the average acceleration between t= 3 s and t= 5 s?
aavg= m/s2 [± 1.4 m/s2]

(E) Consider a straight line that is tangent to the velocity curve at t=1.2 s. You may wish to print the larger version of the graph and use a pencil and ruler to draw the tangent line on the graph.. The slope of that tangent line is the instantaneous acceleration. What is the instantaneous acceleration at t= 1.2 s ?
ax= m/s2 [± 2 m/s2 ]

(F) What is the displacement between t= 0 s and t= 6.5 s?The "area' under the curve, between the two times is the displacement. The "area" is the area enclosed by the curve and the time axis (v=0 line). Those parts of the curve with negative velocity contribute negative area and those with positive velocity contribute positive area.
Between t=0 s and 6.5 s, x = m [± 10 m ]

Explanation / Answer

A) v= 2*62 -12*6 = 0

B) v=-2.875

C)  aavg= v/t = [v(8)-v(0) ]/ (8-0) = 32/8 = 4 m/s2

D) aavg= [v(5) - v(3)]/ (5-3) = 4 m/s2

E) a= dv/dt = 4t-12

a(1.2)= -7.2 m/s2

E) s =(0 to 6.5)vdt = (2t3/3 - 6t2)| = 2/3(6.53) - 6x6.52 = -70.42m

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