Question

a taxi without any passenger moving on a frictionless horizontal Road with a velocity U can be stopped in a distance d.now the passenger add 40% to its weight. what is the stopping distance at velocity U, if the retardation remains the same?

Answer 1

initial velocity =U
final velocity =0
Let the mass of body be m and acceleration be a
Now .
v2= u2- 2as
d = U2/2a ...(1)

Now, force (F) =ma +40ma/100
=140ma/100
=1.4ma =1.4a(m)

Since, acceleration is added so new a = 1.4a

Now , v= 0
u =U
a' = 1.4a

→U2 =2a's
→ s =U2/2a' =U2/(2a ×1.4) =1/1.4 (U2/2a)
= 1/1.4(d) using( 1)