[size=100]A particle moves s along a straight line and passes through a fixed point O. Its velocity, v ms[sup]-1[/sup] is given by v=[math]mt^2+nt[/math], where m and n are constants and t is the time, in seconds, after passing through O. It is given that the particle stops instantaneously when t = 5 s and its acceleration is 3 ms[sup]-2[/sup] when t=1s.[/size][br][size=100][Assume motion to the right is positive][br][br]Find[br][/size]
a) the values of m and of n,[br]
When t=5, v=0,[br] 25m+5n=0[br] 25m=-5n[br] n=-5m[br][br]When t=1,a=3[br] [math]a=\frac{dv}{dt}[/math]= 2mt+n[br] 2m+n=3[br]Substitiute n=-5m into the equation:[br] 2m-5m=3[br] -3m=3[br] m=-1[br] n=-5(-1)=5[br][br]m=-1,n=5
b) the range of values of t when the particle moves to the right.
(-t[sup]2[/sup]+5t)>0[br]t(-t+5)>0[br]t>0 -t+5>0[br] t<5[br]Ans: 0 < t <5
c) The distance, in m, travelled by the particle during the second.
s=[math]\int_1^2-\frac{t^3}{3}-\frac{5t^2}{2}dt=5\frac{1}{6}[/math]m