Manipal Entrance Test 2014 Physics Paper

From given a - t graph acceleration is increasing at constant rate
∴

da

dt

=k( constant )⇒a=kt (by integration)
⇒

dv

dt

=kt⇒dv=ktdt
⇒ ∫dv=k‌∫tdt⇒v=

kt2

2

i.e., v is dependent on time parabolically and parabola is symmetric about v-axis and suddenly acceleration becomes zero, i.e., velocity becomes constant. Hence the most stable graph is (c).