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`        A uniformly accelerated body travels 100m during the 7th second and 110m in the 9th second of its motion. If it is a case of uniform acceleration determine the acceleration and initial velocity. Also calculate the distance traveled by the body during the 11th second of its motion. `
3 months ago

Saurabh Koranglekar
3161 Points
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3 months ago
Aryan
20 Points
```							Hello Dear, questions like these seem tiring and frustrating but in the end it sums up to your basics. let's start from the 1st step,  As uniform acceleration and distance travelled in nth speed is asked ; DŃ= u+a/2 (2n-1) . substitute accordingly .  100=u + a/2{2(7)-1}.  & 110=u+a/2{2(9)-1}.   Using the simultaneous eqns, 10=1/2a(17-13)  . so ur acceleration would be around 5ms^-2. Similarly substitute value of acceleration in any of the 2 given equations of distance (u+a/2 {2(n)-1} to get value of u as 67.5m/s.. Find distance travelled in 11th sec, Dn = 67.5+5/2(2*11-1) = 120m .i hope I mcorrect :)
```
3 months ago
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