Question

1. The "reaction time" of the average automobile driver is about 0.7s . (The reaction time is the interval between the perception of a signal to stop and the application of the brakes.) An automobile can slow down with an acceleration of 10.2ft/s2 .

Part A: Compute the total distance covered in coming to a stop after a signal is observed from an initial velocity of 15.6mi/h . (in a school zone)

Part B Compute the total distance covered in coming to a stop after a signal is observed from an initial velocity of 54.0mi/h .

Answers must be in ft.

2. A rock is thrown vertically upward with a speed of 11.0m/s from the roof of a building that is 50.0mabove the ground. Assume free fall.

Part A In how many seconds after being thrown does the rock strike the ground?

Part B What is the speed of the rock just before it strikes the ground?

3. The rocket-driven sled Sonic Wind No. 2, used for investigating the physiological effects of large accelerations, runs on a straight, level track that is 1080 m long. Starting from rest, it can reach a speed of 1610km/h in 1.60s .

Part A Compute the acceleration in m/s2.

Part B Compute the acceleration in g

Answer 1

1.
15.6 mph = 22.88 ft/s
54 mph = 79.2 ft/s

(a)
time to stop t₁ = v / a = 22.88 ft/s / 10.2 ft/s² = 2.243 s
s = ut + (ut₁ - 1/2 at₁²)
s = 22.88 ft/s * 0.7s + (22.88 ft/s * 2.243s - 1/2 * 10.2 ft/s² * (2.243 s)²
s = 16.016 + (51.32 - 25.66)
s = 41.68 ft

(b)
time to stop, t₁ = 79.2 ft/s / 10.2 ft/s² = 7.765 sec
s = ut + (ut₁ - 1/2 at₁²)
s = 79.2 ft/s * 0.7s + (79.2 ft/s * 7.765s - 1/2 * 10.2 ft/s² * (7.765s)²
s = 55.44 + (615 - 307.5)
s = 362.94 ft

2.
From laws of motion
v^2 = u^2 + 2as............(1)
v = u + at.....................(2)

where v- final velocity, u - initial velocity
a - acceleration due to gravity = g
t - time

applying eqn 1 and putting the respective values, we get
v^2 = 11^2 + 2 * 9.8 * 50
v^2 = 121 + 980 = 1101
v = 33.18 m/s..........answer for (b)

now applying eqn 2 and putting the values, we get
33.18 = 11 + 9.8*t
t = 2.263 sec.......answer for (a)

3.
1610 km/h = 447.22 m/s
(a)
v = u + at
447.22 = 0 + a * 1.6
a = 279.5 m/s²

(b)
a = 279.5 m/s²
a = 279.5 / 9.8 * g m/s²
a = 28.52 g

(c)
v^2 = u^2 + 2as
447.22^2 = 0 + 2 * 279.5 * s
s = 715.6 m

(d)
1020 km/h = 283.33 m/s
v = u + at
0 = 283.33 + a * 1/4
a = -202.38 m/s^2
a = -20.65 g
Hence these figures are not consistent