Math problem regarding kinetic motion

[danong]

Sorry, i have problem in this question which comes from my textbook. I couldn't solve it and i would like someone to guide me.

(Question)
A body travels 30 metres in the sixth second of motion and 24 meters in the ninth second. Find:
i) the initial velocity (ans: 41 m/s )
ii) the retardation, assumed uniform (ans: -2 m/s/s )
iii) the time elapsing before coming to rest (ans: 20.5s)

From my thinking is that, the body is actually moving in a curve so it changes its displacement from 30 meters to 24 meters within the time [6~9] seconds,
but this problem is in charge of kinetic problem as well, as it's in the chapter 'kinetic'.
And since it's not moving in a straightline, i assume that it cannot apply the ordinary equation : S = u * t * 1/2 a*t^2;
But by only reconstructing the equation from a = dv/dt;

Plus, the retardation and initial velocity seems to be dependantly constant variable. So my guess is that, since it involves friction, i would simply guess that it's actually having altering of work as :
Sum(W) = Fn * d(Sn);
where W is the total Work,
Fn is the changing of Forces within the time interval [0,t];
dSn is the delta of distance traveled within time interval [0,t];


Thanks in advance.
Daniel.

 

[CompuChip]

Sorry, i have problem in this question which comes from my textbook. I couldn't solve it and i would like someone to guide me.

(Question)
A body travels 30 metres in the sixth second of motion and 24 meters in the ninth second. Find:
i) the initial velocity (ans: 41 m/s )
ii) the retardation, assumed uniform (ans: -2 m/s/s )
iii) the time elapsing before coming to rest (ans: 20.5s)

From my thinking is that, the body is actually moving in a curve so it changes its displacement from 30 meters to 24 meters within the time [6~9] seconds

From my thinking, this means that the velocity changes from 30 m/s to 24 m/s in the [6, 9] s time interval.

Also, "the retardation is assumed uniform" sounds to me like "the acceleration is constant", which means the formula
s(t) = s_0 + v_0 t + \tfrac12 a t^2
will certainly hold (it follows from solving
s''(t) = a
with the appropriate boundary conditions).

but this problem is in charge of kinetic problem as well, as it's in the chapter 'kinetic'.

"Kinetic" basically just means "movement". E.g. "kinetic energy" is just "movement energy", in a quite literal translation.
But perhaps what is meant is, that you should solve it using energies instead of dynamic equations?

 

[HallsofIvy]

Sorry, i have problem in this question which comes from my textbook. I couldn't solve it and i would like someone to guide me.

(Question)
A body travels 30 metres in the sixth second of motion and 24 meters in the ninth second. Find:
i) the initial velocity (ans: 41 m/s )
ii) the retardation, assumed uniform (ans: -2 m/s/s )
iii) the time elapsing before coming to rest (ans: 20.5s)

From my thinking is that, the body is actually moving in a curve so it changes its displacement from 30 meters to 24 meters within the time [6~9] seconds,
but this problem is in charge of kinetic problem as well, as it's in the chapter 'kinetic'.
And since it's not moving in a straightline, i assume that it cannot apply the ordinary equation : S = u * t * 1/2 a*t^2;
But by only reconstructing the equation from a = dv/dt;

Plus, the retardation and initial velocity seems to be dependantly constant variable. So my guess is that, since it involves friction, i would simply guess that it's actually having altering of work as :
Sum(W) = Fn * d(Sn);
where W is the total Work,
Fn is the changing of Forces within the time interval [0,t];
dSn is the delta of distance traveled within time interval [0,t];


Thanks in advance.
Daniel.

Why do you assert "it's not moving in a straightline"? I can find no where in the problem statement that it is said it is not moving in a straight line. "Kinetic" simply means "moving"- not necessarily moving in a curve.

Assuming constant acceleration, a, (although that's not mentioned until (b)), a body will move (1/2)at²+ v₀t where v₀ is the initial speed. After 5 seconds, it will have gone (25/2)a+ 5v₀. After 6 seconds, it will have gone (36/2)a+ 6v. During the "6^th second", it will have gone (36/2)a+ 6v₀- ((25/2)a+ 5v₀= (11/2)a+ v₀= 30. Do the same for the distances traveled in 8 and 9 minutes, setting the difference equal to 24 and you have two equations to solve for a and v₀.

 

[danong]

oh yes! i misunderstood the question by assuming it has "travelled along". Thanks everyone for the answer and detailed explanation. I got it solved. ^^