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The velocity, Vm/s, of a particle moving in a straight line, at time t seconds is given by
V=1+9/t^2
for 1 is less than or equal to t which is less than or equal to 3. When t=3, the particle is 6m from a fixed point O
on the line.




Find the acceleration when t=2
Find an expression in terms of t for its distance from the fixed point.
Find the distance travelled between t = 1 and 1 = 3.
[4]​

1 Answer

5 votes

Answer:

Acceleration when time is
2 \ sec =
-(5)/(4) \ m/s^2

Distance
s=9-(9)/(t)

Distance travelled between
t=1 and
t=3 is
6\ m

Explanation:

Velocity
V=1+(9)/(t^2) \ \ \ when \ \ 1\leq t\leq 3

Acceleration(a): Rate of change of velocity.


a=(dV)/(dt) =1-(18)/(t^3)

at
t=2


a=1-(18)/(8) \\=1-(9)/(4) \\=-(5)/(4)


a=-(5)/(4) \ m/s^2

Distance(s):


V=(ds)/(dt)\\\\(ds)/(dt)=1+(9)/(t^2)\\\\Integrate\ both\ sides\\\\s=1-(9)/(t)+k\\\\ k\ is\ a\ constant\\\\Given\ s=6\ when\ t=3\\\\6=1-(9)/(3)+k\\\\ k=8\\\\Hence\ s=1-(9)/(t)+8\\\\ s=9-(9)/(t)

Distance between
t=1 \ and \ t=3


=s(3)-s(1)\\\\=(9-(9)/(3))-(9-9) \\\\=6 \ m

User Mfirry
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