Answer
a) velocity at t = 2s will be 6.8m/s
b) Position at t=2s will be 11.8m
Work Step by Step
a) Given acceleration $a = αt$
velocity $v = \int{adt} = \int{αtdt} = \frac{αt^{2}}{2}+ c$
Given $α = 1.2ms^{-3} and c = constant$
Velocity at t = 1s is 5.0 m/s
=> $5 = α/2 + c => c = 4.4 m/s$
To find velocity at t = 2.0s
Using velocity equation and putting t = 2s
$v = \frac{1.2*4}{2} + 4.4 = 6.8 m/s$
b)Position $x = \int{vdt} = \int{(\frac{αt^{2}}{2}+ c)dt} = \frac{αt^{3}}{6}+ ct+d$
Given $x = 6.0m$ at $t=1.0$
=> $6 = \frac{1.2}{6} + c +d$
=> d = 1.4 m
Position at t = 2.0s
$x = \frac{1.2*8}{6} + 4.4*2 + 1.4 = 11.8m$
