Respuesta :
Answer:
a) Maximum speed achieved by Laura = 10.96 m/s and Laura’s acceleration is = 5.54 m/s2
Maximum speed achieved by Helen = 11.33 m/s and Helen’s acceleration is = 3.78 m/s2
b) At 6 the seconds mark Laura was ahead
c) Hence the maximum distance Helen was behind Laura was (20.93m – 15.79m) or 5.14m at 2.89s
Explanation:
To solve this, we require to list out the knowns and the unknowns so as to choose the appropriate equation of motion to work with
Total distance = 100m
Duration of acceleration of Laura = 1.96 s
Duration of acceleration Helen = 3.05 s
Total time to complete the race for both Laura and Helen = 10.4s
Duration of constant speed for Laura = 10.4 – 1.96 = 8.44s
Duration of constant speed for Healn = 10.4 – 3.05 = 7.35s
Maximum speed of Laura = unknown
Maximum speed of Helen = unknown
From the equation of motion
v = u + a × t
Where v is the final velocity,
u is the initial velocity
and t is the time
For the duration of constant acceleration, we have,
For Laura, where aL = Laura’s constant acceleration
u = 0 therefore v = a × t = aL × 1.93 = 1.93aL
For Helen, therefore her final velocity = 3.05aH
Also vL2 = uL2 + 2 × aL × SL1 = 2 × aL ×SL1
And vH2 = uH2 +2 × aH × SH2 = 2 × aH ×SH2
Thus (1.93aL)2 = 2 × aL × SL1 or 3.7249×aL2 = 2 × aL × SL1
or 3.7249aL = 2SL1, also 9.3025aH = 2SH1
but distance at constant speed for Laura = 100 – SL1 = 8.44vL
Thus we have three equations with 3 unknowns and they are
vL2 = 2aLSL1
100 – SL1 = 8.44×vL
VL = 1.93 × aL
solving for SL1 we have 4.65125
3.7249aL = 2SL1
SL1 = 1.86245 aL
or aL = 0.54 SL1, vL = 1.93aL = 1.93× 0.54 S1 = 1.04SL1
and 100 – SL1 = 8.44×vL = 8.44×1.04×SL1 = 8.7SL1
or 100 = SL1 + 8.75SL1 = 9.75×SL1
SL1= 100/9.75 = SL1 = 10.26m, aL = 0.54×10.26 = 5.54 m/s2
vL = 1.93×5.54 = 10.96 m/s
For Healan we have
9.3025aH = 2SH1
vH = 3.05aH
100 - SH1 =7.35×vH again solving for SH1 we have
aH = 0.22SH1, from where vH = 3.05×0.22×SH1 = 0.66SH1 and 100 – SH1 = 7.35×0.66SH1 = 4.82SH1
100 = 5.82SH1 . SH1 = 100/5.825 = 17.17m, vH1 = 0.66SH1 = 11.33m/s and aH1 =0.22SH1 = 3.78 m/s2
The answers are
a) Maximum speed achieved by Laura = 10.96 m/s and Laura’s acceleration is = 5.54 m/s2
Maximum speed achieved by Helen = 11.33 m/s and Helen’s acceleration is = 3.78 m/s2
b) which sprinter is ahead at 6 seconds
for Laura at 6 seconds she is at 10.26 + (6 – 1.96)×10.96 = 54.54m
while Helen is at 17.17+(6 – 3.05) ×11.33 = 50.59m
At 6 the seconds mark Laura was ahead
c) The maximum distance at which Helen was behind Laura can be found by
The maximum distance occurs at the point when Helen speed was equal to Laura’s speed = 10.96m/s
Thus from v = u +at where u = 0 we have 10.96 = 3.78×t, t = 2.89s
At t = 2.89s Laura’s location is S = ut + 0.5at2 =0 + 0.5×5.54×1.962+(2.89-1.96)×10.96 = 10.64 + 10.3 = 20.93m
while that of Helen was 0.5×3.78×2.892 = 15.79m
Hence the maximum distance Helen was behind Laura was (20.93m – 15.79m) or 5.14m at 2.89 s
a) Maximum speed achieved by Laura = 10.96 m/s and Laura’s acceleration is = 5.54 m/s2
Maximum speed achieved by Helen = 11.33 m/s and Helen’s acceleration is = 3.78 m/s2
b) At 6 the seconds mark Laura was ahead
c) Hence the maximum distance Helen was behind Laura was (20.93m – 15.79m) or 5.14m at 2.89s
