- #1
andysmax
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Homework Statement
Two cars are at the start lines of two roads at right angles to each other and pointing to the intersection 500 metres away. Car A starts from rest with uniform accelereation of 0.5m/s2 and accelerates for 20 seconds. It then continues at the speed reached. Car B starts 5 seconds later and 20 metres further away with an acceleration of 0.7m/s2.
Calculate wrt the start time of Car A:
a) Time taken for B and A to be the same distance from the cross-road.
b) The distance from the cross roads at the above time
c) The average velocity of Car A at this distance.
Homework Equations
v=u+at
v2=u2+2as
The Attempt at a Solution
First i used v=u+at and plugged the results into a table showing velocity against time for car A and B, for upto 60 seconds at ten second intervals.
Then i transposed the second equation for s to work out the distance s=v2-u2[tex]/2a[/tex] for car a and s=(v2-u2[tex]/2a[/tex])-20 for car B.
This was then plugged into a second table showing distance against time.
From this table it was deduced that the area of interest was between 50 and 60 seconds.
I carried on with this process until i narrowed it down.
The answer i got for a) was 57.67 seconds.
b) was just working the time back through the previous equations which gave 23.3m (500-476.7).
c) i calculated change in distance over change in time: 476.7/57.67 = 8.267m/s2
However, this was time consuming and i am wondering if there was a simpler way to tackle this(if indeed the above solution is correct!), possibly as a simultaneous equation?
Any help is greatly appreciated.