How to Calculate Distance Traveled in Uniform Acceleration?

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The discussion focuses on calculating the distance traveled by a particle under uniform acceleration using kinematic equations. The key formula derived is Δs(n) = an + u - 0.5a, which represents the distance traveled during the nth second. The problem involves solving a system of equations derived from the distances traveled in the 2nd and 7th seconds, leading to the equations 3a + 2u = 34 and 13a + 2u = 94. By solving these equations, one can determine the values of acceleration (a) and initial speed (u), enabling the calculation of distance for any nth second.

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A particle travels srating with a initial speed u, with uniform acceleration a. Show that the distance traveled during the nth second is u+an-.5a. (b) if the particle travels 17m in the 2nd second of motion and 47m in the 7th second of motion how far will it go in the (i) 10th second of motion (ii) nth second of motion

i worked out the first bit by letting s=un+.5an^2. i then found s=u(n+1)=5a(n+1)^2. and i took these away from each other and i was left with the answer. i no not know what to do for part b though
 
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a) We begin with the kinematic equation:

$$s(n)=\frac{a}{2}n^2+un$$

And so:

$$\Delta s_n=s(n)-s(n-1)=\frac{a}{2}n^2+un-\frac{a}{2}(n-1)^2-u(n-1)=an+u-\frac{a}{2}$$

b) We are given:

$$\Delta s_2=2a+u-\frac{a}{2}=\frac{3}{2}a+u=17$$

$$\Delta s_7=7a+u-\frac{a}{2}=\frac{13}{2}a+u=47$$

Multiplying both equations by 2, we obtain the 2X2 system:

$$3a+2u=34$$

$$13a+2u=94$$

To proceed, subtract the former equation from the latter, eliminating $u$ and solve the result for $a$. Then use either equation to find $u$ using the value you find for $a$. Once you have $a$ and $u$, you will be able to express $\Delta s_n$ in terms of $n$ alone.
 

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