Kinematics problem (uniform acceleration)

Hello,

I've hit rough terrain again, unfortunately. Before I begin I've posted the question I'm stuck on. I've solved part **i** but part **ii** involves part **i**, respectively.

Q: A body starts at and moves with constant acceleration for 4 metres. If it then has a velocity of find:

**i)** how long it will take to move the next 4 m? [A = 0.56 s]

**ii)** how far it will move in the next 3 seconds? [A = 36 m]

For part **i** of the question:

Let , and

Then,

Now, therefore,

From ,

For part **ii** of the question:

Let , then

I don't understand why this does not equate to 36 m? Is the initial velocity for the next 3 seconds ?

Re: Kinematics problem (uniform acceleration)

I thinks you have much easier way to solve the first question,

to find the acceleration, you have the following equation which doesn't need time,

from substitution,

now to find the time that takes the object to move next 4m, use

thus gives,

since t is positive,

I think the part ii is not related with the part i, thus makes the initial velocity

assuming that let's solve the question,

Re: Kinematics problem (uniform acceleration)

Looking back now, I've just misinterupted the question.

I thought part ii, how far it will move in the next 3 seconds, was the add-on on the (v,t) graph after the 8m had been covered. In other words, 8m + 36m = the total distance travelled. 1 s + 0.56 s + 3 s = the total time taken to cover this distance. So to compute the next 3 seconds I thought was the initial velocity for the extra 3 seconds because it was the final velocity after 8m.

Now re-reading the question, **if it then has a velocity of find: part i and part ii.** (Headbang)

Ok no problem, thank you for your efforts (Clapping)