particle moves in a straight line so that t seconds after at a fixed point 0 its velocity Cm/s is given by?

Question

v = t^2 - 5t + 4

a) find the values of t when the particle was at rest
b) find the distance between the positions at which the particle was at rest

please help it's a homework question and i don't know what to do, thank you.

Answer 1

(a) t=1 sec
(b) s=4.5 cm
And this is how:

The general equation for accelerated motion is

S(t)=S0 + V0(t)+a(t^2)/2

S0 – initial distance
V0 – initial velocity
a - acceleration
Do you see where I'm going? :)
However,
There is a trap since it is v(t) = t^2 - 5t + 4
So

S(t)= integral(v(t) dt)=
S(t)= (t^3)/3 – 5(t^2)/2 + 4t + S0

Now let's solve the problem!

(a)The particle was at rest when v(t)=0
We have
t^2 - 5t + 4=0
(t-1)(t-4)=0
t=1 is when the particle was first at rest
(Note at t=4 it will come to rest again.)
(b)I assume they are looking for the distance between t=1 and t=4
Using equation we obtained earlier
S(t)= (t^3)/3 – 5(t^2)/2 + 4t + S0
We have if my math is correct
S(4)=64/3 – 80/2 + 16 + S0
S(1)=1/3 – 5/2 + 4 + S0

S= S(4) - S(1)=63/3-75/2+12= 12 - 99/6 cm
S=4.5 cm

Answer 2

LMAO in basic terms 8 factors eh? communicate approximately one being under favourite! Ya basically can no longer get any understand in any respect, no longer in spite of an amazing answer. LOL Brightest reward, Raji the fairway Witch

Answer 3

(t - 4)(t - 1) = 0
t = 4 or 1 makes this true
b) sorry thats my limit
actually, heres a guess:
dv/dt = 2t-5
distance = (2x4 -5)-(2x1-5) = 3-(-3) = 6
but it is a guess!