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I have made an error somewhere but i'm not quite sure where.

 

Question

A particle travels with uniform acceleration of 4 ms−2 and covers 38 m during the fifth second
of its motion. Using its average velocity in this time period, or otherwise, find its initial speed.

 

My Solution

a = 4 , v = 38 (38/1) t = 5 (5th second)

 

v = u + at

38 = u + 4*5

38 = u + 20

u = 18 ms-1

 

I know I have made an incorrect assumption. What is it and can you explain why my assumption is wrong.

 

Thanks in advance

 Jun 2, 2019
 #1
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\(a=4~m/s^2\\~\\ s_0 = 0\\ s(t) = \dfrac a 2 t^2 + v_0 t = 2t^2 + v_0 t\\ s(6)-s(5)=38~m\\ 2(6^2)+6v_0 - 2(5^2)-5v_0 = 38\\ (72-50)+v_0= 38\\ v_0 = 38-22 = 16~m/s\)

 

If we wanted to use the average speed

 

\(v(t) = v_0 + at\\ \bar{v}=\dfrac{v(6)+v(5)}{2} = \dfrac{11a+2v_0}{2}= \dfrac{11}{2}a+v_0=22+v_0\\ (22+v_0)~m/s\cdot 1~s = 38~m\\ v_0 = 38-22 = 16~m/s\)

.
 Jun 2, 2019

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