Question
A bug is placed on the road. t seconds after it is put down, the function is p(t) = -16tsquared + 64t, where p(t) is measure in feet. At what moment will the bug be 14 feet from where it began?
Answers
visoth
p(t) same as f(x) basically just put in 14 for 7 in the equation but don't put it in for the t in p(t) solve out and that is your answer
p(t)=-16(14)^2 + 64(14)
p(t)=-16(14)^2 + 64(14)
visoth
oh whoops i read the question wrong scratcch that
visoth
oh okay here it is sorry about the above;
p(t)= - 16t^2 +64t
14= - 16t^2 + 64t bring 14 to other side and factor or use the equation -b = or minus square root of (b - 4AC)and divide by 2A and you should get what t is
p(t)= - 16t^2 +64t
14= - 16t^2 + 64t bring 14 to other side and factor or use the equation -b = or minus square root of (b - 4AC)and divide by 2A and you should get what t is
Damon
14= - 16t^2 + 64t
16 t^2 - 64 T + 14 = 0
8 t^2 -32 t + 7 = 0
t = [32 +/-sqrt (1024-224)]/16
= [32 +/- sqrt(800)]/16
= [32 +/- 20sqrt 2]/16
= [ 32 +/- 28.3]/16
= 3.76 on the way down
and
= .231 on the way up
16 t^2 - 64 T + 14 = 0
8 t^2 -32 t + 7 = 0
t = [32 +/-sqrt (1024-224)]/16
= [32 +/- sqrt(800)]/16
= [32 +/- 20sqrt 2]/16
= [ 32 +/- 28.3]/16
= 3.76 on the way down
and
= .231 on the way up