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Rest and Motion: Kinematics — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 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 $= 6.0\ m/s^2)$
Speed $= 54\ km/h$
Braking distance
$a = ...$
Total stopping distance
$b = ...$
Speed $= 72\ km/h$
Braking distance
$c = ...$
Total stopping distance
$d = ...$
$B\ ($deceleration on hard braking $= 7.5\ m/s^2)$
Speed $= 54\ km/h$
Breaking distance
$e = ...$
Total stopping distance
$f = ...$
Speed $72\ km/h$
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 $= 6.0\ m/s^2)$
Speed $= 54\ km/h$
Braking distance
a = 19m
Total stopping distance
$b = 22m$
Speed $= 72\ km/h$
Braking distance
$c = 33m$
Total stopping distance
$d = 39m$
$B\ ($deceleration on hard braking $= 7.5\ m/s^2)$
Speed $= 54\ km/h$
Breaking distance
$e = 15m$
Total stopping distance
$f = 18m$
Speed $72\ km/h$
Braking distance
$g = 27m$
Total stopping distance
$h = 33m$
$\text{a}=\frac{0^2-15^2}{2(-6)}=19\text{m}$
So$, b = 0.2 \times 15 + 19 = 33m$
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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