Asked by joy
can somebody shows me how to get to the answer key step by step. the answer key is 2.78i-10.2j A 1.49-kg particle initially at rest and at the origin of an x-y coordinate system is subjected to a time-dependent force of
F(t) = (4.00ti − 9.00j) N
with t in seconds.What is the particle's total displacement at 12.0 m/s?
Answers
Answered by
Steve
F=ma, so
a(t) = 2.684ti-6.040j
v(t) = 1.342t^2 i - 6.040t j
s(t) = 0.447t^3 i - 3.020t^2 j
Do you mean when the speed is 12.0 m/s? If so, that occurs when
|v|=12.0
√((1.342t^2)^2 + (6.040t)^2) = 12.0
t = 1.839
s(1.839) = 0.447*1.839^3 i - 3.020*1.839^2 j = 2.78i-10.2j
a(t) = 2.684ti-6.040j
v(t) = 1.342t^2 i - 6.040t j
s(t) = 0.447t^3 i - 3.020t^2 j
Do you mean when the speed is 12.0 m/s? If so, that occurs when
|v|=12.0
√((1.342t^2)^2 + (6.040t)^2) = 12.0
t = 1.839
s(1.839) = 0.447*1.839^3 i - 3.020*1.839^2 j = 2.78i-10.2j
Answered by
joy
what don't get it. ecan you explain more on the part √((1.342t^2)^2 + (6.040t)^2) = 12.0
t = 1.839 ?
t = 1.839 ?
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