/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 A body is thrown up in a lift wi... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 11

A body is thrown up in a lift with an upward velocity \(u\) relative to the lift from its floor and the time of flight is found to be \(t\). The acceleration of the lift will be (A) \(\frac{u-g t}{2}\) (B) \(\frac{u+g t}{2}\) (C) \(\frac{2 u-g t}{t}\) (D) \(\frac{u}{t}-g\)

Short Answer

Expert verified
The acceleration of the lift is given by: \[a = \frac{u}{t}-g\]

Step by step solution

01

Set up the given information

We know that the body is thrown up in the lift with an upward velocity \(u\) relative to the lift and the time of flight is \(t\).
02

Write the equations of motion for the body relative to the lift

The upward motion of the body relative to the lift is given by: \[h_{up} = ut - \frac{1}{2}gt^2 \] Where \(h_{up}\) is the height reached by the body relative to the lift during the upward motion.
03

Write the equations of motion for the body during the downward motion

The downward motion of the body relative to the lift is given by: \[h_{down} = \frac{1}{2}gt^2\] Where \(h_{down}\) is the distance the body falls relative to the lift during the downward motion.
04

Combine the equations in steps 2 and 3 to find an expression for the total motion of the body relative to lift

Since the body eventually lands on the ground, the total height reached by the body relative to the lift is the sum of the heights in the upward and downward motions: \[h_{total} = h_{up} + h_{down}\] Substitute the equations from steps 2 and 3 and we get: \[h_{total} = ut - \frac{1}{2}gt^2+ \frac{1}{2}gt^2\] Which simplifies to: \[h_{total} = ut\]
05

Write the equation of motion for the lift

Since lift moves upwards with an unknown acceleration \(a\), we need to find the lift's height \(h_{lift}\) as a function of time during the flight time \(t\) using an equation of motion. We will use a slightly modified version of the Kinematic equation that doesn't require the final velocity: \[h_{lift} = a\frac{t^2}{2}\]
06

Write the equations of motion for the body relative to the observer on the ground

When considering the observer on the ground, the height reached by the body is the sum of the height reached by the body relative to the lift and the height reached by the lift: \[h_{total} + h_{lift} = ut + a\frac{t^2}{2}\] But since the body eventually comes back to the initial height relative to the ground, we have: \[h_{total} + h_{lift} = 0\]
07

Solve for the acceleration of the lift

Plug the equation for \(h_{total}\) from step 4 into the equation for total height and solve for \(a\): \[ut - a\frac{t^2}{2}=0\] Solving for \(a\), we get: \[a=\frac{2u}{t}-g\] This matches the option (D). Thus, the acceleration of the lift is given by: \[a = \frac{u}{t}-g\]

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