Question

An
object starting from rest travels 20 m in first 2 s and 160 m in next 4 s. What will be the velocity after 7 s from the start.

Answer 1

from 2nd eq of motion
s=ut^2+1/2 at^2
u=0 ,s=20,t=2sec
20=1/2a2×2
we got a=10m/s^2
velocity at the end of 2 sec=
v=u+at
v=at. (u=0)
v=10×2
v=20m/s

no need of second part of 4 sec as we have to only calculate velocity after 7 sec from start
as u=0
a=10 (from above)
by 1st eq of motion
v=u+at
v=at
v=10×7
v=70m/s
so velocity after 7 sec from start is 70m/s

Answer 2

here,
u = 0
$v_{1} = \frac{20}{2} = 10 \: \: m {s}^{ - 1}$
$v_{2} = \frac{160}{4} = 40 \: \: m {s}^{ - 1}$
therefore,
$a = \frac{v_{2} - v_{1}}{t} = \frac{40 - 10}{6} = \frac{30}{6} = 5 \: \: m {s}^{ - 2}$
so now, t = 7 seconds
applying first equation of motion,
$v_{3} = v_{3} + at \\ = > v_{3} = 10 + (5)(7) \\ = > v_{3} = 10 + 35 \\ = > v_{3} = 45 \: \: m {s}^{ - 1}$
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