Question

tmysicstecture Lecture 902021Homework+2.1.pdf 5- Egg Drop. You are on the roof of the physics building, 46.0 m above the ground (Fig P2.80). Your physics professor, who is 1.80 m tall, is walking alongside the building at a con- stant speed of 1.20 m/s. If you wish to drop an egg on your pro fessor's head, where should the professor be when you release u=1.20 m/s the egg? Assume that the egg is in free fall 130D D00E 0300 0030 46.0 m 0102 6- Entering the Freeway. A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. The driver accelerates with constant acceleration along the ramp and onto the freeway The car starts from rest, moves in a straight line, and has a speed of 20 m/s (45 mi/h) when it reaches the end of the 120-m-long ramp. (a) What is the acceleration of the car? (b) How much time does it take the car to travel the length of the ramp? (c) The traffic on the freeway is moving at a constant speed of 20 m/s. What dis tance does the traffic travel while the car is moving the length of the ramp?

can you explain how to solve 5 and 6, I'm struggling and don't understand

Answer 1

(5) Time taken by the egg to reach on the professor's head which will be given as -

y = y₀ + v₀ t + (1/2) g t²

where, y = height of the physics building = 46 m

y₀ = height of physics professor = 1.80 m

v₀ = initial speed of an egg = 0 m/s

g = acceleration due to gravity = 9.8 m/s²

then, we get

(46 m) = (1.80 m) + (0 m/s) t + (0.5) (9.8 m/s²) t²

[(46 m) - (1.80 m)] = (4.9 m/s²) t²

t² = [(44.2 m) / (4.9 m/s²)]

t = _$\sqrt{}$9.02 s²

t = 3 s

Distance traveled by the physics professor in this time which will be given as -

x = v_i t

where, v_i = initial speed of physics professor = 1.20 m/s

then, we get

x = [(1.20 m/s) (3 s)]

x = 3.60 m

So, the physics professor should be 3.60 m away from the bottom of building when we released the egg.

(6) A car sits in an entrance ramp to a freeway, waiting for a break in the traffic.

a. What is the acceleration of a car?

using equation of motion (3), we have

v² = v₀² + 2 a $\Delta$x

(20 m/s)² = (0 m/s)² + 2 a (120 m)

a = [(400 m²/s²) / (240 m)]

a = 1.67 m/s²

b. How much time does it take the car to travel the length of ramp?

using equation of motion (1), we have

v = v₀ + a t

(20 m/s) = (0 m/s) + (1.67 m/s²) t

t = [(20 m/s) / (1.67 m/s²)]

t = 11.9 s

t $\approx$ 12 sec

c. What distance does the traffic travel while a car is moving the length of ramp?

we know that, x = v t = [(20 m/s) (12 s)]

x = 240 m