Velocity (m/sec) 6- Figure at right shows the velocity-time graphs for objocts A
ID: 1880540 • Letter: V
Question
Velocity (m/sec) 6- Figure at right shows the velocity-time graphs for objocts A and B again. a. What is the situation of the position and velocity of both objects at 7.5 sec? The positions for objects A and B are isxA-10m and xp-15 m when t-0 sec. 10 b. Is the magnitude of the acceleration of object A greater than, less than or equal to the magnitude of the acceleration of object B? Explain your Time (se lis c. Calculate the displacement and distance traveled by both objectsI during the first 8 seconds. d. Is the magnitude of the acceleration of object A greater than, less than or equal to the magnitude of the acceleration of object B exactly at t-10 sec? Explain your reasoning. e. Which object is moving to left at t-12 sec? 150 100 f. Graph the position-time for both particles. 50 ime (ec) -50Explanation / Answer
The graph basically shows how the velocities of objects A and B varies from t=0sec. It could be observed that the the Va is a straight line going all the way from 20m/s at t=0sec to 0 velocity at t=10sec and further. Basically the velocity of object A had been constantly reducing and the same of object B has been constantly increasing from -10m/s at t=0sec to 10m/s at t=10sec and beyond.
The rate of change of velocities are constant i.e, the slopes of the lines are both 2 each in magnitude.
We could therefore write Va(t) = -2t+20 (because slope is -2 and intercept value on V axis is 20)
Vb(t)=2t-10.
We can substitue t=7.5sec above and see that values of velocities at 7.5sec are Va(7.5) = 5m/s and Vb(t) = 5m/s. This is indicated in the graph by a meeting point of both lines.
since acceleration is constant, distance x is given by
x=ut+at2/2
=20*7.5 - 2*7.52/2
=93.75m.
The final position of object A at t=7.5sec is therefore 10(initial position)+93.75 = 103.75m.
But B moves in negative direction till t=5sec and from then on reverses the direction and accelerates forward.
distance travalled by B till 5sec = -10*5+2*52/2 = -25m. This gives position as 15(initial)-25 = -10m at t=5sec.
From t=5sec to 7.5 sec, it accelerates and travels a distance of 0*2.5+2*2.52/2 = 6.25.
Therefore the final position of B at t=7.5 sec is -3.75m.
The magnitudes of acceleration of both objects is same because as seen from the graphs, their slopes are same. For instance consider t=0 to t=1 interval. A's velocity drops from 20m/s to 18m/s i.e., 2 m/s2. During the same interval B's velocity increases from -10m/s to -8m/s i.e., 2m/s2 again. Just that A is decelerating and A is accelerating but both with same magnitude.
Object A's distance till 8sec = 20*8-2*82/2=96m. The distance and displacement of A both are same because motion of A has been only in singe direction during 8 secs.
Object B's distance till 8sec = -10*8+2*82/2 = 16m. The displacement till 5 sec is -25m as calculated in part a). From t=5sec to t=8sec, x=2*32/2 = 9m. Hence the final positition of B at t=8sec is 15(initial)-25(t=5)+9(t=5 to 8) = -1 and displacement therefore is (intital - final) which is 15 - (-1) = 16m.
The magnitudes of acceleration of both A and B are same throughout and hence same at t=10sec as well.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.