Enhanced EOC: Exercise 3.13 A car comes to a bridge during a storm and finds the

Question

Enhanced EOC: Exercise 3.13 A car comes to a bridge during a storm and finds the bridge washed out. The driver must to the other side, so he decides to try leaping it with his car. The side the car is on is 20.9m above the river, whereas the opposite side is a mere 2.4m above the river. The river itself is a raging torrent 56.0 m wide. You may want to review (DA-pages 77-85) For help with math skills, you may want to review: Vector Magnitudes For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Different initial and final heights. Part A How fast should the car be traveling just as it leaves the cliff in order to just clear the river and land safely on the opposite side? mm/S Submit Hints My Answers Give Up Review Part Incorrect; Try Again; 2 attempts remaining Part B What is the speed of the car just before it lands safely on the other side? mm/S

Explanation / Answer

Using the equation of motion,

s = ut +0.5*gt^2

In vertical motion,

u = initial vertical speed = 0 <---- dives horizontally

g = 9.8 m/2

s = vertical displacemnt = 20.9 - 2.4 = 18.5 m

So, 18.5 = 0 + 0.5*9.8*t^2

So, t = 1.94 s

In horizontal motion,

Using the equation,

s = ut +0.5*at^2

Now, u =initial horizontal speed

a = 0 <--- no force horizontally

s = horizontal displacement = width of river = 56 m

So, 56 = u*t + 0

So, u = 56/t = 56/1.94 = 28.8 m/s <--------answer

B)

Using the equation of motion,

v = u + at

In vertical direction,

v = 0 + 9.8*1.94 = 19.01 m/s <----- vertical speed at the end of the bridge

So, speed of the car at the other end = sqrt(v^2+u^2)

= sqrt(19.01^2+28.8^2)

= 34.5 m/s <-------answer

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