Recent content by mcr

So I proceeded to use the equation: 12.6(t-2)+.5a(t-2)^2=198 ft. and solved for (t-2) in terms of a. I then substituted back into the equation and I get a=.0663 ft/sec^2. This is not the answer, so I must have done something incorrect. Does anything stand out that is wrong? Thanks.

Ok, this is what I got for the area under the curves which represents their distances. For the faster runner I got d=14.7t. For the slower runner I got d=12.6t+1/2a(t-2)^2. So now the two distances are equal so 14.7t=12.6t+1/2a(t-2)^2. Is this correct? I have one equation with two unknowns...

So the slope is the acceleration, but what about b?

Thanks, I finally got #1. On to #2. The velocity vs time graph for one runner is going to be a horizontal line. I found it to be y=14.67. The velocity vs time graph for the other runner has two pieces. For t<=2 I have y=12.57, but for t>2 it is a slanted line (y=mx+b), but I don't know how...

ok, I'm needing more assistance. problem 1: I'm assuming the minimum total velocity is at the peak of its flight. Is this correct? But, I'm still stuck. problem 2: for the fast runner I have d=14.67(ft/sec)t+4.2 and for the other runner I have d=1/2at^2. I'm not sure on these...

Hi, I'm new at this and I need some help with the following practice problems: 1) At what angle relative to the horizontal must an object be launched if its minimum velocity in flight is 28% of its launch velocity? The answer is to be in degrees. I have the answer to be 73.7. 2)...