Physics - Kinematics Question with Solution | TestHub

Let $u$ be initial velocity and $a$ be acceleration.

Distance in $n$-th second: $S_n=u+a(n-\frac{1}{2})$.

$S_3=u+a(3-\frac{1}{2})=u+\frac{5a}{2}=4$ m (1)

$S_5=u+a(5-\frac{1}{2})=u+\frac{9a}{2}=12$ m (2)

Subtracting (1) from (2): $2a=8⟹a=4$ m/s².

Substitute $a=4$ into (1): $u+\frac{5(4)}{2}=4⟹u+10=4⟹u=-6$ m/s.

Distance in next 3 seconds (6th, 7th, 8th): $S_6+S_7+S_8$.

$S_6=-6+4(6-\frac{1}{2})=-6+4(\frac{11}{2})=-6+22=16$ m.

$S_7=-6+4(7-\frac{1}{2})=-6+4(\frac{13}{2})=-6+26=20$ m.

$S_8=-6+4(8-\frac{1}{2})=-6+4(\frac{15}{2})=-6+30=24$ m.

Total distance $=16+20+24=60$ m.