1. A particle P moves in a straight line so that its displacement, s metres, from a fixed point O is given by s=4+15t-t(cube) where t is the time in seconds after passing though a point X on the line.
a) Find the distance OX.
b) Find the velocity of the particle at t=2.
c) Calculate the acceleration of the particle when it is momentarily at rest.
d) Calculate the distance moved by the particle during the third second of its motion.
2. The volume V cm(cube) in a container of height h cm is given by
V=pie/2 (216h-h(cube)). Given that h is changing at a rate of 1/2 cm/s, find the rate of change of V when h = 6cm, giving your answer in terms of pie.
Hi,
For Q1, the following facts are important and should be memorised:
Velocity v = ds/dt,
Acceleration a = dv/dt,
Displacement = integral of v with respect to t,
Velocity = integral of a with respect to t.
In (a), consider t = 0.
In (c), what is v when the particle is momentarily at rest.
For Q2, we are given that
V is an expression involving h,
dh/dt = 1/2,
h = 6
and we want to find dV/dt.
Applying the chain rule, we have dV/dt = (dh/dt) (dV/dh), where the value of dV/dh can be found with h = 6.
Try it, thanks!
Cheers,
Wen Shih
i solved question 1 before u reply, and i think i had the correct answer.
attempting qns 2 atm :D thks alot !
finish qns 2, ans is 36pie right? :D
for qns 1,
a) 4m.
b) 3m/s
c) -13.4m/s(square)
d)4m