ID webassign.net | Hw3 - Ch 2 (Motion in One Dimensio) Webassign.net Need Help?

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ID webassign.net | Hw3 - Ch 2 (Motion in One Dimensio) Webassign.net Need Help? Read It 5. -2 Points 8. 1/3 points | Previous Answers SerCP11 2.P.035 My Notes Ask Your Speedy Sue, driving at 34.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 175 m ahead traveling at 5.40 m/s. Sue applies her brakes but can accelerate only at-1.80 m/s2 because the road is wet. Will there be a collision? Yes No If yes, determine how far into the tunnel and at what time the collision occurs. If no, determine the distance of closest approach between Sue's car and the van. f no, ente" for the time.) distance TI time Your response differs from the correct answer by more than 100%, s Need Help?Read It Submit Answer Save ogress Practice Another Version 9· -/6 points SerCP102.AE.010. My Notes Ask Your EXAMPLE 2.10 A Rocket Goes Ballistio GOAL Solve a problem involving a powered ascent followed by free-fall motion. Maximum height - PROBLEM A rocket moves straight upward, starting from rest with an acceleration of +29.4 m/s2. It runs out of fuel at the end of 4.00 s and continues to coast upward, reaching a maximum height before falling back to Earth. (a) Find the rocket's velocity and position at the end of 4.00 s. (b) Find the maximum height the rocket reaches. (c) Find the velocity the instant before the rocket crashes on the ground. Phase 2 Rocket fuel STRATEGY Take y = 0 at the launch point and y positive upward, as in the fiqure. The problem consists

Explanation / Answer

for sue:

initial velocity vo = 34 m/s

initial position x0 = 0

acceleration a = -1.8 m/s^2

for van :

x0 = 175 m

v0 = 5.4 m/s

a = 0

for sue :x = 34*t+1/2(-1.8)*t^2

for van: x-175 = 5.4*t

x = 175+5.4t

if both positons are same

34t-t^2 = 175+5.4t

t^2-28.6t+175 = 0

solving for t

t = 8.87 or 19.7 sec

here t will be small so t = 8.87 sec

x = 34*8.87-8.87^2

x = 222.9 m

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