Question

while moving at 40 m/s, an object accelerates uniformly by 6 m/s2 in 9 sec.
a. what is the velocity after this time interval ?
b. how far does it go during this period?

Answer 1

$v = u + at \\ u = 40 \\ a = 6 \\ t = 9 \\ hence \: velocity \: after \: 9 \: seconds \: \\ is \: v = 40 + 6 \times 9 = 94 \\ s = ut + a {t}^{2} \div 2 \\ hence \: distance \: travelled \: in \: this \\ interval \: is \\ s = 40 \times 9 + 6 \times 9 \times 9 \div 2 = 603m$
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