The re truck is moving forward at a speed of 35 mi /hr and is decelerating at the rate of 10 ft/sec2. Simultaneously, the ladder is being raised and extended. At the instant considered the angle is 30° and is increasing at the constant rate of 10 deg /sec. Also at this instant the extension b of the ladder is 5 ft, with b ˙ = 2 ft/sec and b ¨ =−1 ft/sec2. For this instant determine the acceleration of the end A of the ladder (a) with respect to the truck and (b) with respect to the ground.

(a). -1.76i + 0.76j ft/s^2.

(b). -10.42i + 5.7j ft/s^2.

Explanation:

So, we are given the following data or parameters in the question/problem above;

=> "The fire truck is moving forward at a speed = 35 mi /hr and is decelerating at the rate = 10 ft/sec2."

=> "At the instant considered the angle = 30° and is increasing at the constant rate = 10 deg /sec. Also at this instant the extension b of the ladder = 5 ft, with b ˙ = 2 ft/sec and b ¨ =−1 ft/sec2"

STEP ONE: Convert the angular speed to rad/s and speed to ft/s.

Angular speed => π/18 rad/s

Speed => 35 × 1.46667 = 52.33 ft/s

STEP TWO: determine the coordinate form of the acceleration of the truck.

= - 10(cos 30°)i - sin(30°)j [π/18 × K] × 25i[π/18 × K] × 2i × [π/9 × K] - 1i.

= -10.42i + 5.7j ft/s^2.

PLEASE NOTE: The step TWO above is the solution to the second part (b) of this question which is the with respect to the ground.

STEP THREE:

This, from step two above;

( -10.42i + 5.7j ) - ( acceleration of the truck).

= - 10.42i + 5.7j - - 10(cos 30°)i - sin(30°)j = -1.76i + 0.76j ft/s^2.

answer:

a

explanation:

answer:

explanation:

assault rifle is a rapid-fire, magazine-fed rifle designed for military use. it is a shoulder-fired weapon that allows the shooter to select between semi-automatic (requiring you pull the trigger for each shot), fully automatic (hold the trigger and the gun continuously fires) or three-shot-burst modes.