The graph above shows the velocity of an object vs time. (A) What is the (instan

Question

The graph above shows the velocity of an object vs time.

(A) What is the (instantaneous) velocity of the object at t= 0.5 s?
v(0.5 s)= ______m/s [± 0.2 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= 3 s?
v( 3 s)=  m/s. [± 0.2 m/s ]

(C) What is the velocity of the object at t= 4.5 s?
v( 4.5 s)= ______m/s. [± 0.2 m/s ]

(D) What is the average acceleration between t= 0 s and t= 5 s?
aavg=  m/s2 [± 0.2 m/s2 ]

(E) What is the average acceleration between t= 2 s and t= 5 s?
aavg= ______m/s2 [± 0.2 m/s2 ]

(F) What is the displacement between t= 1 s and t= 5 s?
?x= _____m [± 0.54 m ]

(G) What is the displacement between t= 0 s and t= 4 s?
?x= ______m [± 0.56 m ]

Explanation / Answer

(A) At t = 0.5 s, v = 3.1 m/s

(B) At t= 3 s , v = 4 m/s

(C) At t= 4.5s , v = 2.5 m/s

(D) Average acceleration = ( change in velocity in time interval)/ time interval = (velocity at 5s - velocity at 0s)/(5-0)

= (1-4)/5 = -0.6 m/s2

(E) Average acceleration = ( change in velocity in time interval)/ time interval = (velocity at 5s - velocity at 2s)/(5-2)

= (1-4)/3 = - 1 m/s2

(F) Displacement between 1s and 5 s = Area under the graph in this time interval = Area of triangle BCL + Area of BLIJ + area of CDHI + Area of trapezium DFGH = 0.5x(4-2)(2-1) + 2(2-1) +(4-2)4 + 0.(1+4)(5-4) = 13.5 m

(G) Displacement between 0s and 4 s = Area under the graph in this time interval = Area of trapezium ABJK + Area of triangle BCL + Area of BLIJ + area of CDHI = 0.(4+2)(1) + 0.5x(4-2)(2-1) + 2(2-1) +(4-2)4 = 14 m

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.