A body possessing zero initial velocity covers a distance 27 m during the 5th second of its motion.Calculate the acceleration of the body .

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Solution

Step 1:Given that,
Initial velocity (u) =0.
Distance (s)=27 m.
nth second=5
a = acceleration
Step 2:Formula.
$S_{n} = u + \frac{a}{2} (2 n - 1) .$

Step 3:Calculation.
$S_{n} = u + \frac{a}{2} (2 n - 1) .$
$27 = 0 + \frac{a}{2} (2 (5) - 1)$ .
$2 \times 27 = a (9) . a = \frac{54}{9} . a = 6 \frac{m}{s^{2}} .$

Hence,
Acceleration of the body (a) = 6 $\frac{m}{s^{2}} .$ .

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A body possessing zero initial velocity covers a distance 27 m during the 5th second of its motion.Calculate the acceleration of the body .

Step 1:Given that,Initial velocity (u) =0.Distance (s)=27 m.nth second=5 a = accelerationStep 2:Formula.Sn=u+a2(2n-1). Step 3:Calculation.Sn=u+a2(2n-1).27=0+a2(2(5)-1).2×27=a(9).a=549.a=6ms2. Hence,Acceleration of the body (a) = 6ms2..