Answer to Question #95073 in Classical Mechanics for Yogendra pandey

Question

(a) You and your new alien friend, Howard need to escape area 51 fast! Luckily, Howard knows of a high-end sports car you can use in your escape. Assuming you don't get shot in the process, what is the minimum acceleration necessary to reach a linear displacement of 5.00 km in under 2.00 minutes? Assume you start from rest and maintain a constant acceleration. (b) Howard's Bugatti has a maximum acceleration around 11.0 m/s2 and a top speed of about 260.0 mph. What is the least amount of time it will take you to reach a linear displacement of 5.00 km if you accelerate at a constant rate of 11.0 m/s until reaching top speed and then cruise at the top speed for the remainder of the trip into Mexico? (

Answer 1

a)Using the law for uniformly accelerated motion we can express a travelled distance in time $t$ (and we use the condition that we start from the rest ${v_0} = 0$)

$$l = {v_0}t + \frac{{a{t^2}}}{2} = \frac{{a{t^2}}}{2}$$

Expressing acceleration

$$a = 2\frac{l}{{{t^2}}}$$

We get minimal value of acceleration needed to reach the given displacement if we give a travelled distance is equal to the given displacement. Now do the calculations

$${a_{\min }} = 2\frac{l}{{{t^2}}} = 2\frac{{5[{\text{km}}]}}{{{2^2}[{{\min }^2}]}} = 2\frac{{5 \cdot {{10}^3}[{\text{m}}]}}{{{2^2} \cdot {{60}^2}[{{\text{s}}^2}]}} = \frac{{25}}{{36}}[\frac{{\text{m}}}{{{{\text{s}}^2}}}] \approx 0.7[\frac{{\text{m}}}{{{{\text{s}}^2}}}]$$

b)At the first let's evaluate the time needed to reach the top speed from the rest (denote it by $\tau$) . Use the law for uniformly accelerated motion

$$v = {v_0} + a\tau = a\tau$$ $$\tau= \frac{v}{a}$$

We will need to convert mph in m/s, it's can be done by multiplying by ${0.447}$. Thus

$$\tau = \frac{{{v_{\max }}}}{a} = \frac{{260[{\text{mph]}}}}{{11[\frac{{\text{m}}}{{{{\text{s}}^{\text{2}}}}}]}} \approx \frac{{0.447 \cdot 260[\frac{{\text{m}}}{{\text{s}}}{\text{]}}}}{{11[\frac{{\text{m}}}{{{{\text{s}}^{\text{2}}}}}]}} \approx 10.566[{\text{s}}]$$

In this time the car travelled

$$d = \frac{{a{t^2}}}{2} = \frac{1}{2} \cdot 11[\frac{{\text{m}}}{{{{\text{s}}^{\text{2}}}}}] \cdot {10.566^2}[{{\text{s}}^2}] = 614.022[{\text{m}}]$$

The rest part of the full distance $x = l - d$ the car will move with constant (top) speed. It will take the time

$$t = \frac{{l - d}}{{{v_{\max }}}} = \frac{{5[{\text{km}}] - 614.022[{\text{m}}]}}{{260[{\text{mph]}}}} \approx \frac{{5 \cdot {{10}^3}[{\text{m}}] - 614.022[{\text{m}}]}}{{0.447 \cdot 260[\frac{{\text{m}}}{{\text{s}}}{\text{]}}}} \approx 37.739[{\text{s}}]$$

Thus the full time to reach the given displacement is

$${t_{{\text{full}}}} = t + \tau = 10.566[{\text{s}}] + 37.739[{\text{s}}] = 48.305[{\text{s}}] \approx 48.3[{\text{s}}]$$

This is the answer to part b)