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Rest and Motion: Kinematics — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsRest and Motion: Kinematics4 Marks
Question
Complete the following table:
Car Model
Driver X
Reaction time 0.20s
Driver Y
Reaction time 0.30s
A (deceleration on hard braking = $6.0m/s^2$)
Speed = 54km/h
Braking distance
a = ...
Total stopping distance
b = ...
Speed = 72km/h
Braking distance
c = ...
Total stopping distance
d = ...
B (deceleration on hard braking = $7.5m/s^2$)
Speed = 54km/h
Breaking distance
e = ...
Total stopping distance
f = ...
Speed 72km/h
Braking distance
g = ...
Total stopping distance
h = ...
✓

Answer

Car Model
Driver X
Reaction time 0.25s
Driver Y
Reaction time 0.35s
A (deceleration on hard braking = $6.0m/s^2$)
Speed = 54km/h
Braking distance
a = 19m
Total stopping distance
b = 22m
Speed = 72km/h
Braking distance
c = 33m
Total stopping distance
d = 39m
B (deceleration on hard braking = $7.5m/s^2$)
Speed = 54km/h
Breaking distance
e = 15m
Total stopping distance
f = 18m
Speed 72km/h
Braking distance
g = 27m
Total stopping distance
h = 33m
$\text{a}=\frac{0^2-15^2}{2(-6)}=19\text{m}$
So, b = 0.2 × 15 + 19 = 33m
Similarly other can be calculated.
Braking distance: Distance travelled when brakes are applied.
Total stopping distance = Braking distance + distance travelled in reaction time.

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