ch02-p037 - zero which means there is only one root. The...

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37. (a) We note that v A = 12/6 = 2 m/s (with two significant figures understood). Therefore, with an initial x value of 20 m, car A will be at x = 28 m when t = 4 s. This must be the value of x for car B at that time; we use Eq. 2-15: 28 m = (12 m/s) t + 1 2 a B t 2 where t = 4.0 s . This yields a B = – 2.5 m/s 2 . (b) The question is: using the value obtained for a B in part (a), are there other values of t (besides t = 4 s) such that x A = x B ? The requirement is 20 + 2 t = 12 t + 1 2 a B t 2 where a B = –5/2. There are two distinct roots unless the discriminant 10 2 2( 20)( a B ) is zero. In our case, it is
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Unformatted text preview: zero which means there is only one root. The cars are side by side only once at t = 4 s. (c) A sketch is shown below. It consists of a straight line ( x A ) tangent to a parabola ( x B ) at t = 4. (d) We only care about real roots, which means 10 2 2( 20)( a B ) 0. If | a B | > 5/2 then there are no (real) solutions to the equation; the cars are never side by side. (e) Here we have 10 2 2( 20)( a B ) > 0 two real roots. The cars are side by side at two different times....
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