Question

somebody plz answer this, irrelevant answers will be reported. A body moving with a uniform velocity 4m/s undergoes uniform acceleration 5m/s2.find distance travelled in 10sec and distance travelled in 10th sec.(distance travelled in a particular second or distance travelled in 'n' th second=Sn=u+n-1/2a​

Answer 1

Answer:

$u_{1}$=10 m/s

Explanation:

Sn = distance travelled in n sec-distance travelled in (n-1) sec

Sn =  ($u_{1} n + \frac{1}{2} an^{2}$) - ($u_{1} ( n-1) + \frac{1}{2} a(n-1)^{2}$

    =  $u_{1} +\frac{1}{2} a (n^{2} -(n-1)^{2} )$

    =  $u_{1} + a_{n} -\frac{a}{2}$

On comparing it with:  $s_{n} = 0.4 + 9.8$

      =  $u_{1} -\frac{0.4}{2} = 9.8$

      =  $u_{1} = 10 m/s$

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