Question

15 when breaks are applied on a moving
vehicle, it stops after travelling a distance.
This distance is called stopping distance.
Write an expression for stopping
distance in terms of initial velocity
(u) and retardation (a).
If the initial speed is doubled keeping the
retardation same, by how much will!
the stopping distance change?
[March-2011)​

Answer 1

Answer:

Explanation:

Velocity of moving vehicle=v  

0

​  

 and decceleration =a

Final velocity of the vehicale =0 (As it stops)

Let the stopping distance be s

Then applying the third equation of motion we have:

v  

2

=v  

0

2

​  

+2(−a)s

⇒0=v  

0

2

​  

−2as

⇒s=  

2a

v  

0

2

​  

 

​  

→ Expression for stopping distance.

Answer 2

Answer:

Velocity of moving vehicle=v0 and decceleration =a

Final velocity of the vehicale =0 (As it stops)

Let the stopping distance be s

Then applying the third equation of motion we have:

v2=v02+2(−a)s

⇒0=v02−2as

⇒s=2av02→ Expression for stopping distance.

Explanation:

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