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Honors Physics Motion In One Dimension HW #3 Complete the following problems on a separate sheet of paper. Significant Figures are to be used.

1. The skid marks left by the decelerating jet-powered car The Spirit of America were 9.60 km long. If the car’s acceleration was -2.00 m/s2, what was the car’s initial velocity?

∆𝑥 = 9.60  𝑘𝑚 = 9.60  𝑥  10!𝑚

𝑎 = −2.00𝑚𝑠!

𝑣! =  0

𝑣!! =  𝑣!! + 2𝑎∆𝑥

𝑣!! =  −2𝑎∆𝑥

 𝑣! =   −2𝑎∆𝑥 =   −2   −2.00  𝑚𝑠!

9.60  𝑥  10!𝑚 = 196𝑚𝑠

2. The black mamba is one of the world’s most poisonous snakes, and with a maximum speed of 18.0 km/h, it is also the fastest. Suppose a mamba waiting in a hideout sees pretty and begins slithering towards it with a velocity of +18.0 km/h. After 2.50 s, the mamba realizes that its prey can move faster than it can. The snake then turns around and slowly returns to its hideout in 12.0 s. Calculate:

a) The mamba’s average velocity during its return to the hideout. 18.0  𝑘𝑚

ℎ    𝑥  

1  ℎ𝑜𝑢𝑟3600  𝑠𝑒𝑐𝑜𝑛𝑑𝑠

 𝑥  1000  𝑚1  𝑘𝑚  

= 5.00𝑚𝑠

𝑠 =𝑑𝑡

𝑑 = 𝑠𝑡 = 5.00𝑚𝑠∗ 2.50  𝑠 = 12.5  𝑚

𝑣 =  ∆𝑥∆𝑡

=  12.5  𝑚12.0  𝑠

= 1.04𝑚𝑠

b) The mamba’s average velocity for the complete trip.

𝑣 =  ∆𝑥∆𝑡

=  0  𝑚14.5  𝑠

= 0𝑚𝑠

c) The mamba’s average speed for the complete trip.

𝑠 =𝑑𝑡=  12.5  𝑚 + 12.5  𝑚

14.5  𝑠= 1.72

𝑚𝑠

3. In a scientific test conducted in Arizona, a special cannon called HARP (High Altitude Research Project) shot a projectile straight up to an altitude of 180.0 km. If the projectile’s initial speed was 3.00 km/s, how long did it take the projectile to reach its maximum height?

𝑣! = 0,  

𝑣! = 3.00𝑘𝑚𝑠

∆𝑦 = 180.0  𝑘𝑚

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∆𝑦 =12𝑣! +  𝑣! ∆𝑡

∆𝑡 =  ∆𝑦

12 (𝑣! +  𝑣!)

=  180.0  𝑘𝑚

12 0 + 3.00 𝑘𝑚𝑠

= 120  𝑠

4. In 1934, the wind speed on Mt. Washington in New Hampshire reached a record high. Suppose a very sturdy glider is launched in this wind, so that in 45.0 s the glider reaches the speed of the wind. If the glider undergoes constant acceleration of 2.29 m/s2, what is the wind’s speed? Assume that the glider is initially at rest.

𝑣!   = 0

𝑎 = 2.29  𝑚𝑠!

∆𝑡 = 45.0  𝑠

𝑣! =  𝑣! + 𝑎∆𝑡 = 0 + 2.29  𝑚𝑠!

45.0  𝑠 =  103𝑚𝑠

5. If you throw a ball straight upward, it will rise into the air and then fall back down toward the ground. Imagine that you throw the ball with an initial velocity of 13.7 m/s.

a) How long does it take the ball to reach the top of its motion? 𝑣! =  0

𝑣! =  13.7𝑚𝑠

𝑎 =  −9.8𝑚𝑠!

𝑣! =  𝑣! + 𝑎∆𝑡 𝑣! = 𝑣! + 𝑎∆𝑡

∆𝑡 =  −𝑣!𝑎

=  −13.7𝑚𝑠−9.8  𝑚𝑠!

= 1.40𝑚𝑠

b) How far will the ball rise before it begins to fall? 𝑣!! =  𝑣!! + 2𝑎∆𝑦

𝑣! =  0

𝑣! =  13.7𝑚𝑠

𝑎 =  −9.8  𝑚𝑠!

∆𝑦 =?

∆𝑦 =  𝑣!! −  𝑣!!

2𝑎=  (0)! − (13.7𝑚𝑠 )

!

2  (−9.8  𝑚𝑠!)= 9.58  𝑚  (𝑀𝑎𝑘𝑒𝑠  𝑠𝑒𝑛𝑠𝑒𝑠  𝑡ℎ𝑎𝑡  𝑖𝑡  𝑖𝑠  𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒  𝑏𝑒𝑐𝑎𝑢𝑠𝑒  𝑖𝑡  𝑖𝑠  𝑚𝑜𝑣𝑖𝑛𝑔  𝑈𝑃!)

c) What is its average velocity during this period?

𝑣!"# =  𝑣! +  𝑣!

2=  0 + 13.7𝑚𝑠

2= 6.85

𝑚𝑠

6. Anna walks off the end of a 10.0-m diving platform. a) What is her acceleration in m/s2 toward the pool?

𝑎 =  −9.8  𝑚𝑠!

b) How long does it take her to reach the water?

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∆𝑦 =  𝑣!∆𝑡 +12𝑎∆𝑡!, 𝑣! = 0

∆𝑡 =  2∆𝑦𝑎   =  

2 ∗ (−10.0  𝑚)

−9.8  𝑚𝑠!= 1.43  𝑠𝑒𝑐𝑜𝑛𝑑𝑠

c) What is her velocity when she reaches the water?

𝑣! =  𝑣! + 𝑎∆𝑡, 𝑣! = 0

𝑣! =  𝑎∆𝑡 = −9.8  𝑚𝑠!

1.43  𝑠𝑒𝑐 =  −14.0𝑚𝑠

7. A rocket used to lift a satellite into orbit undergoes a constant acceleration of 6.25 m/s2. When the rocket reaches an altitude of 45 km above the surface of Earth, it is traveling at a velocity of 625 m/s. How long does it take for the rocket to reach this speed?

𝑣! =  𝑣! + 𝑎∆𝑡, 𝑣! = 0

∆𝑡 =  𝑣!𝑎=625𝑚𝑠6.25𝑚𝑠!

= 1.00  𝑥  10!  𝑠𝑒𝑐𝑜𝑛𝑑𝑠

8. On the surface of Mars, free-fall acceleration is 0.379 times as much as that on the surface of Earth. A robot on Mars pushes a rock over a 500.0-m cliff.

a) How long does it take the rock to reach the ground below the cliff?

𝑎 = 1.379   −9.8𝑚𝑠!

=  −13.5𝑚𝑠!

∆𝑦 =  𝑣!∆𝑡 +12𝑎∆𝑡!

𝑣! = 0

∆𝑡 =  2∆𝑦𝑎   =  

2  (−500.0  𝑚)

−13.5  𝑚𝑠!= 8.61  𝑠𝑒𝑐𝑜𝑛𝑑𝑠

b) How fast is the rock traveling when it reaches the surface?

𝑣! =  𝑣! + 𝑎∆𝑡, 𝑣! = 0

𝑣! =  𝑎∆𝑡 = −13.5  𝑚𝑠!

8.61  𝑠𝑒𝑐 =  −116𝑚𝑠

c) How long would it take the rock to fall the same distance on the surface of Earth?

∆𝑦 =  𝑣!∆𝑡 +12𝑎∆𝑡!

𝑣! = 0

𝑎 =  −9.8  𝑚𝑠!

∆𝑡 =  2∆𝑦𝑎   =  

2(−500.0  𝑚)

−9.8  𝑚𝑠!= 10.1  𝑠𝑒𝑐𝑜𝑛𝑑𝑠

9. A sky diver jumps from an airplane 1000.0 m above the ground. He waits for 8.0 s and then opens his parachute. How far above the ground is the sky diver when he opens his parachute?

∆𝑦 =  𝑣!∆𝑡 +12𝑎∆𝑡!, 𝑣! = 0

∆𝑦 =  0 +12−9.8

𝑚𝑠!

(8.0𝑠)! =  −313.6  𝑚

𝑎𝑏𝑜𝑣𝑒  𝑔𝑟𝑜𝑢𝑛𝑑 = 1000.0  𝑚 − 313.6  𝑚 = 686.4  𝑚