Question

A particle is moving with uniform acceleration. In the 11th and 15th seconds from the
beginning, it travels 7.2 m and 9.6 m respectively. Find its initial velocity and acceleration
with which it moves?
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Answer 1

Answer :-

We know that the distance travelled in nth second is given by ,

$\sf S_n = u +\dfrac{1}{2}a [2n-1]$

Distance travelled in 11th sec .

=> S_11 = u + ½a [2(11)-1]

,=> S_11 = u + ½a [22-1]

=> S_11 = u + 21a/2

=> 7.2 m = u + 21a/2

Distance travelled in 15th sec .

=> => S_15 = u + ½a [2(15)-1]

,=> S_15 = u + ½a [30-1]

=> S_15 = u + 29 a/2

=> 9.6 m = u + 29a/2

S_15 - S_11 :-

=> 9.6 - 7.2 = 29a/2-21a/2

=> 2.4 = 8a/2

=> a = 2.4 /4

=> acclⁿ = 0.6 m/s²

Put this in first case ,

=> 7.2m = u + 21 * 0.6 /2

=> 7.2 m = u + 21*0.3

=> 7.2 m = u + 6.3

=> u = 7.2 - 6.3

=> u = 0.9 m/s

Answer 2

The formula to find the distance travelled in nth second is ,

S_n = u + 1/2 a [2n - 1 ]

First case ,

s = 7.2 m

=> 7.2 = u + 1/2 a [2(11)-1]

=> 7.2 = u + 1/2a (22-1)

=> 7.2 = u + 21a/2

Second case ,

=> 9.6 = u + 1/2 a [2(15)-1]

=> 9.6 = u + 1/2a (3p-1)

=> 9.6 = u + 29a/2

Subtract both ,

=> 9.6 -7.2 = 29a/2 - 21a/2

=> 2.4 = 4a

=> a = 0.6 m/s²

Substitute this value ,

=> 9.6 = u + 29*0.6 /2

=> 9.6 = u + 29*0.3

=> 9.6 = u + 8.7

=> u = 0.9 m/s

Therefore the acceleration is 0.6m/s² and initial velocity is 0.9 m/s