Question

10) Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a large toy rocket to the back of a sled and take the modified sled to a large, flat, snowy field. You ignite the rocket and observe that the sled accelerates from rest in the forward direction at a rate of 12.5 m/s² for a time period of 3.50 s. After this time period, the rocket engine abruptly shuts off, and the sled subsequently undergoes a constant backward acceleration due to friction of 4.15 m/s².

After the rocket turns off, how much time does it take for the sled to come to a stop?

time:———

By the time the sled finally comes to a rest, how far has it traveled from its starting point?

distance traveled:———-

14) A small coin, initially at rest, begins falling. If the clock starts when the coin begins to fall, what is the magnitude of the coin's displacement between ?1=0.219 s and ?2=0.984 s?

Δ?=

15)

You are an engineer in charge of designing a new generation of elevators for a prospective upgrade to the Empire State Building. Before the state legislature votes on funding for the project, they would like you to prepare a report on the benefits of upgrading the elevators. One of the numbers that they have requested is the time it will take the elevator to go from the ground floor to the 102nd floor observatory. They are unlikely to approve the project unless the new elevators make the trip much faster than the old elevators.

If state law mandates that elevators cannot accelerate more than 1.50 m/s2 or travel faster than 12.3 m/s, what is the minimum time in which an elevator can travel the 373 m from the ground floor to the observatory floor?

minimum travel time:——-

Answer 1

10)

Given,

The time period, t = 3.50 sec

The acceleration, a = 12.5 m/s²

The velocity gained by the sled during the time of acceleration, v = at

v = 12.5 * 3.50

= 43.75 m/s

Acceleration produced, a = 4.15 m/s²

a)

The time taken by the sled to come to a stop, t = v / a

t = 43.75 / 4.15

= 10.54 sec

b)

The distance travelled during the first 3.50 sec is d₁ = (1/2) * at²

d₁ = (1/2) * 12.5 * (3.50)²

= 76.56 m

The distance travelled during the last 10.54 sec is, d₂ = vt - (1/2) * at²

d₂ = 43.75 * 10.54 - 0.5 * 4.15 * (10.54)²

= 461.125 - 230.52

= 230.61 m

The total distance, d = d₁ + d₂

d = 76.56 + 230.61

d = 307.17 m