MHB 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, starting with the kinematic equation for distance. The formula for distance traveled during the nth second is derived as u + an - 0.5a. For part b, the distances traveled in the 2nd and 7th seconds are used to set up a system of equations to solve for the initial speed (u) and acceleration (a). By manipulating these equations, the values of u and a can be determined, allowing for the calculation of distance traveled in the 10th and nth seconds. The process emphasizes the importance of correctly applying kinematic principles to solve for unknowns in motion problems.
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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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