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How can I work out the following:- If a car is stationary and then accelerates to 20m.p.h., what distance will it cover in 1 second, 2 seconds etc. whilst it is still accelerating? The car has a 0-60 mph time of 12 seconds. Will be great if you can help. Thanks. (I would have asked the U.K. science line, but they have run out of funds!)
Answer 1:

Any mechanics text book will tell you more about your question, but I will give you some guidelines which will help you with the calculations..
The change of velocity (dv), with time (dt), gives you the acceleration (a), of the car.
So a=dv/dt
You also need to check for the units:
60mph = (1/60)mile/sec, so the car goes from 0 to 1/60mile/sec in 12sec.
Therefore, a=1/12sec*1/60mile/sec=1/720(mile/sec^2).
The distance (x), after time (t) is x=v0t+0.5at*t.
v0 is the initial velocity, which is 0 in this case.
So x=0.5*1/720=(1/1440)miles after 1 sec,
And x=0.5*4/720=(1/360)miles after 2 sec.
And so on.

Answer 2:

To answer this problem, one must first recognize that distance traveled is the integral of velocity, and velocity is the integral of acceleration. Therefore,the distance traveled can be calculated as the area under the curve describing the relationship between velocity and time. (Conversely, acceleration is the derivative of velocity, and velocity is the derivative of distance. A derivative can be interpreted as the rate of change of a parameter, so that velocity is the rate of change of distance, and can be calculated as the slopeof the relationship between distance and time.)

In order to calculate the distances traveled at times 1 and 2 seconds, the velocity must be converted from miles/hour to miles/second using the following conversion factors:

60 miles/hour * 1 hour/60 minutes * 1 minute/60 seconds = 1/60 miles/second

If you assume that the car has a constant acceleration from 0 to 1/60miles/second in 12 seconds, then you can graphically depict this acceleration as a linear plot of velocity versus time, where the velocity, v, at time t = 0seconds is
v = 0 miles/second, and the velocity at time t = 12 seconds is
v =1/60 miles/second.

The relationship between velocity and time is therefore the line
v = a*t, where v is velocity in miles/second, t is time in seconds, and a is the acceleration in miles/(second) 2, or the slope of this line. The acceleration is the rate of change, or derivative, of velocity. Since you already know two points on this line, you can calculate the acceleration or slope of the line. The two points (v, t) on the line are (0 miles/second, 0 seconds) and (1/60miles/second, 12 seconds). Therefore, 1/60 miles/second = a*12 seconds, and acceleration,
a = 1/60 miles/second * 1/12 seconds = 1/720 miles/(second) 2.

The relationship between velocity and seconds is described as
v = [1/720miles/second2]*t.
As described earlier, distance is merely the integral of velocity (since velocity is the derivative of distance). If you do some basic calculus and take the anti-derivative, or integral, of this equation, then you find that
d = *(1/720 miles/second)*t2 (so that d = v).

Therefore, at 1 second, the distance traveled is
*(1/720 miles/second)*(1second) 2 = 1/1440 miles. At 2 seconds, the distance traveled is
*(1/720miles/second)*(2 second) 2 = 4/1440 miles = 1/360 miles.


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