Question

( u=0 )
constant arceleration for 6 seconds
( s=u t+frac{1}{2} a t^{2} quad d_{1}+alpha_{2}=frac{1}{2} a(4)^{2} )
( S=frac{1}{2} a times(2)^{2} )
( S=2 a=d_{1} quad d_{1}+d_{2}=frac{1}{2} a cdot 168 )
[
begin{array}{rl}d_{1}=2 a & d_{1}+d_{2}=8 a d_{2}=6 aend{array}
]

# U V 2 1 U1 +U2 ut uz 22 9) A particle starts from rest and experiences a constant acceleration for 6 seconds. If it travels a distance d, in the first two seconds, a distance dy in next two seconds

Solution