Question

An object is moving along a coordinate line subject to acceleration a(t)=(2t+3)cm/s^2
If v(0)=-4 and s(0)=0
determine:
a.)the total distance traveled on [0,4] secs
b.)displacement on [0,4]sec

please help!

Answer 1

hey man it's 1AM here, I gtg to sleep, could be please the work and answer, I'd high appreciate it! I'll look at it first thing in the morning, thank you very much! PLEASE do leave work and answer, both are important, thank you very much! i'll medal and fan in morning, promise, thanks bts

Answer 2

could you*** please leave the work and answer here

Answer 3

You can integrate once to get the velocity function
\[v(t)=\int\limits 2t+3 dt=t^2+3t+c\]
now we need to find a integration constant that satiefies v(0)=-4
\[v(0)=0^2+3*0+C=-4\]
So the velocity function for this problem is
\[v(t) = t^2+3t-4\]
then we integrate again to get the distance function
\[s(t)=\int\limits t^2+3t-4dt=\frac{ t^3 }{ 3 }+\frac{ 3 }{ 2 }t^2-4t +C\]
Then we again have to find the integration constant
\[s(0)=\frac{ 0^3 }{ 3 }+\frac{ 3 }{ 2 }0^2-0t +C=0\]
so the distance function is
\[s(t)=\frac{ t^3 }{ 3 }+\frac{ 3 }{ 2 }t^2-4t +C\]
now just plug in 4s to find the distance

Answer 4

thank you very much, sorry for late reply, I got it now @bts30