1. To drive to Chicago, a distance of ahout 150 km in 2 hours, you must average

Question

1. To drive to Chicago, a distance of ahout 150 km in 2 hours, you must average a speed of 75 km/hr. However, due to traftic leaving Milwaukee, you can only average a speed of 40 km/hr for the first half-hour. How fast must you drive to keep an average speed of 75 km/hr 2. Aaron Rodgers steals Usain Bolt's gold medal and takes off running at t-0 accelerating at m/. At t-2 s, Usain notices his missing gold medal and takes off after Aaron Rodgers accelerating at 2m/s2.When does Usain catch Aaron Rodgers? At what speed does Usain tackle Aaron? 3. A bicyclists accelerales from a stop to 10 mls in about 15 & What is the average acceleration of cyclist? 4. Suppose the bicyclist above rolls down a hill with velocity r, = (10.0 + 7.01) mk. (a) what is the instantancous velocity of the cyclist at 5s? (b) What is the average acceleration of the cyclist between 1 sand 5 s? (c) what is the instantaneous acceleration at t = 5 s? 5. Suppose you drop a rock from a height of 30 m. Over the last second of its fall, how far does the rock travel? 6. You drop a rock from a height h the same moment your friend throws a rock upward at speed v.when the rocks collide, beh rocks are moving at opposite directions, but the peed of the falling rock is twice that of the thrown rock. what heightkao. did the collision occur? 7. Suppose I tell you that the acceleration of a particle is given by a(t)-41+1ms.a) find Rs velocity as a function of time ifatt=0,ithad a velocity of-2ms (bPind its position as a function of time if it starts out atx

Explanation / Answer

6) let the dropped rock is A and vertically projected rock is B

given that speed of falling rock A = speed of thrown rock B

VA = 2*VB

let at a height h_collide,they are colliding each other

distance travelled by A before colliding is (h-h_collide)

Speed of A at collision is VA = sqrt(2*g*(h-h_collide))

speed of B at the collision is VB

using kinematic equations

VB^2 -v^2 = 2*a*h_collide
a = -g

VB^2 = v^2-2g*h_collide

VB = sqrt(V^2-2g*h_collide)

VA = 2*VB

sqrt(2*g*(h-h_collide)) = 2*sqrt(V^2-2g*h_collide)

2*g*(h-h_collide) = 4*(V^2-2*g*h_collide)

6*g*h_collide = 4V^2 -2*g*h

h_collide = (4V^2-2*g*h)/(6g)

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