Question

The motion of a particle along a straight line is governed by the relation a= t^3-2t+7 where a is in m/s^2 and t is time in seconds. At time t=1 the velocity of the particle is 3.58m/s and displacement is 9.39m. Calculate the displacement,velocity and acc at t=2s.


I know the formula but the calculation part is getting wrong somewhere.

Answer 1

you have to integrate that function in order to find the velocity v(t):
\[\Large v\left( t \right) = \int {\left( {{t^3} - 2t + 7} \right)dt} \]

Answer 2

v = \[\rm v=\frac{ t^4 }{ 4 } -2\frac{ t^3 }{ 3 }+7t+C\]

Answer 3

brb

Answer 4

ok! now you have to substitute your condition, namely t=1 and v=3.58, so you will find the value of the constant C

Answer 5

I got this:
\[\Large v\left( t \right) = \int {\left( {{t^3} - 2t + 7} \right)dt} = \frac{{{t^4}}}{4} - {t^2} + 7t + C\]

Answer 6

yes

Answer 7

after a substitution, you will get this:
\[\Large 3.58 = \frac{1}{4} - 1 + 7 + C\]

Answer 8

\[\sf 3.58=\frac{ 1 }{ 4 }-\frac{ 2 }{ 3 } +7+C\]

Answer 9

why 2/3?

Answer 10

OHHHHHHHHHHHHHHHHHHhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
i got it correctly now because i copied the question wrong

Answer 11

is a(t) like this:
\[\Large a\left( t \right) = {t^3} - 2t + 7\]