Using MATLAB: The velocity, v, and the distance, d, as a function of time, of a car...

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Mechanical Engineering

Using MATLAB:

The velocity, v, and the distance, d, as a function of time, ofa car that accelerates from rest at constant acceleration, a, aregiven by

= a n d = 12

Determine v and d at every second for the first 10 seconds for acar with acceleration of = 15 ft/s2. Your output must have exactlythe same format as the template table below. Note that dots havebeen added to the table below; you can count the dots to determinethe exact spacings. Also note the units.

ร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทTime (s) ร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยท ร‚ยทร‚ยทร‚ยทDistance (ft) ร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทVelocity(mph)ร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทx.xร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทx.xxe+yyร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทร‚ยทx.xxx

Answer & Explanation Solved by verified expert
3.7 Ratings (299 Votes)

MATLAB CODE

clear
clc;
a = 15; %ft/s2 (acceleration)
for i = 1:10
ร‚ย ร‚ย ร‚ย  t(i) = i; %time span
ร‚ย ร‚ย ร‚ย  d(i) = 12 + (1/2)*a*t(i)^2; % initial distane is 12 ft.
ร‚ย ร‚ย ร‚ย  v(i) = a*t(i)*(3600/5280); % 1 mile = 5280 ft and 1h =3600 sec
end
table = [t' d' v']

Ans

Time(s) Distance(ft) velocity (mph)
1.0000ร‚ย ร‚ย ร‚ย ร‚ย ร‚ย  19.5000 10.2273
2.0000ร‚ย ร‚ย ร‚ย  42.0000 20.4545
3.0000ร‚ย ร‚ย ร‚ย ร‚ย ร‚ย  79.5000 30.6818
4.0000ร‚ย ร‚ย ร‚ย ร‚ย  132.0000 40.9091
5.0000ร‚ย ร‚ย ร‚ย  199.5000 51.1364
6.0000ร‚ย ร‚ย ร‚ย  282.0000 61.3636
7.0000ร‚ย ร‚ย ร‚ย ร‚ย  379.5000 71.5909
8.0000ร‚ย ร‚ย ร‚ย ร‚ย  492.0000 81.8182
9.0000ร‚ย ร‚ย ร‚ย ร‚ย  619.5000 92.0455
10.0000ร‚ย ร‚ย ร‚ย  762.0000 102.2727

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