An object travels 14 cm in fourth second of motion

[9giddjl]

Homework Statement

An object starts from rest and travels 14 cm in the fourth second of its motion. Calculate:
a) the acceleration of the object (assumed uniform)
b) the distance traveled by the object after 4.0 s.
c) the distance traveled in the tenth seconf of motion, if the acceleration is the same throughout its motion.

Homework Equations

u=0 s=0.14 m t= 1 sec?

The Attempt at a Solution

I really don't get these questions.. the ones that are 'it travels 2 metres in 5th second of motion:S however if it traveled 14 cm in fourth second of motion does that mean the time would just be one second? For question a) could i use s=ut +0.5at^2? Wait scratch that i just tried it and it didn't work.. umm pls help. These type of questions just confuse me altogether. Thankyou very much

 

[alphysicist]

Hi 9giddgl,

If it travels 14cm in the fourth second of motion, that means that it means it travels 14cm from t=3 seconds to t=4 seconds.

You do want to use the equation you mentioned ($\Delta x = v_0 t + \frac{1}{2}at^2$). But you can use it twice, to find the position at 3 and 4 seconds (not as a number, but as an expression). What do you get for these two positions?

Once you have those, how can you use the 14cm talked about in the problem to find the acceleration?

 

[9giddjl]

So, for the two positions, is it s(3)=4.5a and s(4)=8a?
and then s(4)-s(3) = 14cm = 8a - 4.5a = 3.5a.
Therefore the answer is 4cm/s^2 or 0.04m/s^2?

Is this right, because the answer at the back of my question sheet said 4m/s^2?

Thanks

 

[alphysicist]

The book must have meant 4 cm/s^2.

If the acceleration had been 4 m/s^2, then after one second it would be moving 4 m/s; after two seconds 8 m/s; at three it would be going 12 m/s, which would mean at the start of the fourth second it would be going 1200 cm/s. Since it only does 14 cm in that second, those numbers just don't match.