Page 191 - Physics

Linear motion

come to a stop. The fi nal velocity will Example 8.8
become zero, hence, v = 0. Now, applying
the third equation of motion in this case A car has a velocity of 20 m/s and the
of non-uniform motion: braking capacity of the car is given as
2
v  2 u  2 2as 10 m/s . What will the stopping distance
be for the car?
0 u 2  2as
Solution:
u 2
s =− Since the acceleration occurs in
2a negative direction, a = -10 m/s 2
Now, it is important to notice that a is Applying formula for stopping distance,
taken in negative since the vehicle is u 2
decelerating. Hence, the stopping distance s =− 2a
will fi nally come out to be positive.
  2
20
Deriving formula for Reaction time s   2  10 

Imagine the case of a free-falling object
and the time it takes to react and catch it  400
to fi nd the reaction time. Keep in mind s  20
that the initial velocity will be zero since
the object is freely falling. Applying the s  20 m
second equation of motion to fi nd the
reaction time of the vehicle: Example 8.9
1
s  ut  at 2 How will the stopping distance be
2 affected if the velocity of the vehicle is

u  0 m/s doubled?

Solution:
1
s  gt 2 Imagine in the fi rst case, the velocity
2 of the vehicle was v m/sec and the
acceleration was a m/s . Now, stopping
2
Where g is acceleration due to gravity.
distance in the fi rst case is given as,
1 2
s  (10 t 2 ) s =− u
2 2a
2s 2
t = v 
2
10 s  2a
1
2
t  0.20s In the second case, when the velocity
is doubled, v   2 m/sv , acceleration =
Therefore, t  0.20s a m/s .
2
Stopping distance is given as,

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Physics Form 1 Final.indd 185 16/10/2024 20:58