Question

a body start from rest and with a uniform acceleration of 10 m/s2 for 5 sec. during the next 10 second with uniform speed. the total distance traveled
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Answer 1

Answer:

Distance covered in first 5$4sec{ s }_{ 1 }=0*5+\dfrac { 1 }{ 2 } *10*{ 5 }^{ 2 }\\ =125 m$$

Distance covered in last 10 sec s

2

=vt=(u+a∗5)t

=(0+10∗5)10=500m

Total distance. D=s

1

+s

2

=625m