Question

An aircraft attains a speed of 175 mi/h in 2.5 s. (a) What is acceleration? (b) Distance traveled?

Answer 1

you must change 175 mil/h -> m/s and find acceleration

Answer 2

yep, its simple

Answer 3

but this is physics, right ?

Answer 4

Since you start with a rate in miles per hour but have time in seconds, you need to change units. How many seconds are in an hour? That's easy, there are 60 seconds in a minute, and 60 minutes in an hour. So first convert "miles per hour" to "miles per second". Average acceleration is velocity (speed) divided by time. Then the distance traveled can be found with d=rt, where d is distance, r is rate of travel, and t is time. This is called the "dirt equation" in American English.

Answer 5

summary of "erflynn" response:

Conversion is needed SI Unit for acceleration is \[m/\sec^2\]
\[a=v/t\]a = accel., v= velocity (speed), t=time \[d=rt\]d = distance, r = rate of travel, t = time

-------------
Testing response/input system for equations. Still needs a lot of work. I can't edit a post though.This open study is a great system.

Answer 6

a=dv/dt
dv=v in this question but you must change it to the unit of m/s
since 1 mile = 1609.344 m 1 hr = 3600 sec then.
A m/s = 175 * (1609.344/3600)=175*0.44704=78.232 m/s
a=78.232/2.5=31.
Since velocity-time graphic is linear,
Distance d=(v*t)/2=(78.232*2.5)/2=97.79 m

Answer 7

wouldn't the distance traveled be d=641.667 ft?
you don't have to change anything to meters just get it to common terms. Miles, feet, hour, sec.
and the acceleration be a=252,016.129 mph?

Answer 8

This is simple algebra, perhaps 7th or 8th grade (with some minimal physics that tell you how to interpret the units and how velocity, time and acceleration are related); not sure how this is related to a calculus course. Is the question in the course?

Since the question doesn't specify what units you need to use, I'm guessing you can use whatever units you prefer. In which case you should prefer metric units (IMHO).

You also need to assume that the aircraft started at rest (since the question is asked imprecisely; it could in principle have attained 175 mph having been at 170 before).

And another necessary assumption: The aircraft used constant acceleration to attain its current speed (highly unrealistic, especially when starting from 0). Then the graph of speed vs. time will be a straight line, and the distance travelled will be (top_speed/time_taken)/2, since the distance is the area under the curve (the area makes a triangle, which has half the area of the rectangle).

Using simply speed * time for distance would only work at constant speed.

So erflynn's "dirt equation" doesn't apply in this case; the answer should be half that, d=rt/2.

Answer 9

(okay, the "area under the curve" is calculus, admittedly)