The next three questions concern the following situation. A pickup truck starts

Question

The next three questions concern the following situation. A pickup truck starts from rest and maintains a constant acceleration a0. After a time t0, the truck is moving with speed 25 m/s at a distance of 60 m from its starting point. When the truck has travelled for a time t1 = t0/2, its speed is v1. Which of the following statements concerning v1 is true? v1 12.5 m/s When the truck has travelled a distance of 30m, its speed is v2. Which of the following statements concerning v2 is true? V2 12.5 m/s How long does it take for the pickup to reach its speed of 25 m/s? t0 = 3.1 s t0 = 4.8 S t0 = 6.8 S t0 = 9.6 S t0 = 13.4 S This question and the next one concern the physical situation shown below. Two balls are are released from a table top of height H = 3m at the same time. The green ball is given an initial velocity vG = 4 m/s in the downward direction, while the red ball is given an initial velocity vR = 8 m/s in the upward direction. Compare v bottom(rec), velocity with which the red ball strikes the floor with v bottom(green), the velocity with which the green ball strikes the floor. v bottom(red) v bottom(green) How long does it take for the green ball to reach the floor? t(green) = 0.35 s t(green) = 0.47 s t(green) = 1.29 s t(green) = 1.79 s t(green) = 1.95 s John leaves Urbana at noon and travels due east at an unknown constant speed. One hour later, Meg leaves from a distance 270 miles due east of Urbana and travels at a constant speed of 60 mph in a due west direction. John and Meg pass each other 3 hours after John has left Urbana. What was John's speed? VJ = 40 mph VJ = 50 mph VJ = 70 mph VJ = 90 MPH VJ = 135 mph

Explanation / Answer

(1). v = a*t =>25 = a*t S = 0.5*a*(t^2)=>60 = 0.5*a*(t^2)=>120 = a*(t^2) 120/25 = a*(t^2)/(a*t) =>t = 4.8 s a = 25/4.8 m/s^2 v1 = a*t/2 = 25/2 = 12.5 m/s Option (B) (2) S = 0.5*25*t^2/4.8 S = 30 => 30*2*4.8/25 = t^2 => t = 3.394 s v2 = a*t = 25*3.394/4.8 = 17.68 m/s Option (C) (3) Obtained in (1) t0 = 4.8s Option (B) (4) Net height attained by red ball => 3 + v^2/(2*g)=>3 + 3.2 = 6.2 m Let g = 10m/s Final velocity of redball,Vr => sqrt(2*g*6.2) =11.136 m/s Final velocity of ball,Vg => sqrt(2*g*3 + 4^2) =8.718 m/s Vr>Vg Option (C) (5) S = 0.5*g*(t^2) 6.2 = 0.5*10*t^2 =>t^2 = 1.24 => t1 = 1.114 s (time of descent) time of ascent,t2 => initial velocity/g = 0.8 sec Total time = t1+t2 = 1.92 sec Option(E) (6) Distance travelled by John = 270 - 2*60mph = 150 miles Speed of John = 150/3 = 50mph Option (B)

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