Answer:
Step-by-step explanation:
Given a body starts with a velocity of 15 cm/s and an acceleration of 4 cm/s², you want (1) the distance covered in the 6th second, and (2) the distance covered in the 7th and 8th seconds.
The speed is the sum of the initial velocity and the increase due to acceleration:
v = 15 +4t . . . . . cm/s after t seconds
The distance is the integral of the velocity.
[tex]\displaystyle s =\int_{t_0}^{t_1}{(15 +4t)}\,dt=15(t_1-t_0)+2(t_1^2-t_0^2)\\\\s=(t_1-t_0)(2(t_1+t_0)+15)[/tex]
The sixth second is the one ending at t=6, so the distance is ...
s = (6 -5)(2(6 +5) +15) = 1(2(11) +15) = 37
The distance covered in the 6th second is 37 cm.
This is the time period between t=6 and t=8, so the distance is ...
s = (8 -6)(2(8 +6) +15) = 2(2(14) +15) = 86
The distance covered in the 7th and 8th seconds is 86 cm.
