Initial velocity, u=0,a=1ms–2,t=10s , If Sn be the distance covered by the three-wheeler in nth second. Then
Sn=u+

a

2

(2n−1),
Sn=

a

2

(2n−1)=

1

2

(2n−1)
For n=1,
S1=

1

2

(2×1−1)=0.5 m
n=2,
S2=

1

2

(2×2−1)=1.5 m
n=3,
S3=

1

2

(2×3−1)=2.5 m and so on
The following is the table for the distance Snth travelled in nth second.Also Dn= distance covered by the particle in 10 sec.
Dn=un+

1

2

an2
=0×10+ 12×1×102
(n=10)
=50 m.
As the equation describing the relation between Dn and n is of seconddegree, so the graph of the vehicle from start to 10‌thsecond is expectedto be a parabola. Velocity of the vehicle at the end of 10‌th second isv=0+1×10=10ms–1 and it will move with this velocity after 10 seconds.Thus the graph will be a straight line inclined to time axis for uniformlyaccelerated motion. The plot of the distance covered by the three wheelerwith time is shown in the graph.