M3 Question before tomorrow please

Can someone please show me how to do this question:

A particle P moves in a straight line with acceleration (6t)ms² at time t seconds. Its displacement from a fixed point O fo the line at this time is x metres. Given that x=6 when t=1 and x=75 when t=4 find an expression for x in terms of t.

Reply 1

freakeystyley34

right

I think this should be right

a = 6tms^-2

when x=6 t=1

when x=75 t=4

a=dv/dt

so dv/dt=6t

so v=integral of 6t dt = 3t^2+c

v=dx/dt

so dx/dt=3t^2+c

x=integral of 3t^2+c dt = t^3+ct+k

you can use the data they gave you to work out c and t with simultaneous equations.

Hope that was clear.

Reply 2

john!!

a = 6t
v = 3t^2 + u
x = t^3 + ut + s

6 = u + s
75 = 64 + 4u + s

75 = 70 + 3u
3u = 5
u = 5/3

s = 10/3

x = t^3 + 5t/3 + 10/3

its just using the given information to find the integration constants