Answer:
Laura is ahead and for a distance of 3.22 m
Explanation:
To solve this problem of one-dimensional kinematics, we have to find the acceleration and the final speed of each runner. Let's start with Laura
Lura data is acceleration time 1.82s, total run time 10.4 s and total distance 100m. In all the races the rest starts, so the initial speed is zero (Vo = 0)
Vf1= Vo + a1 t1
Vf1 = x/t
XT = X1 + X2
X1 = Vo t1 + ½ a1 t1²
X1 = ½ a1 t1²
X2 = Vf1 (t-t1)
This is the remaining time of the race after the acceleration is over.
XT = ½ a1 t1² + Vf1 (t-t1)
We remplace the expression of Vf1
XT = ½ a1 t1² + a1 t1 (t-t1)
Laura's aceleration (a1) is
a1= XT / [ ½ t1² + t1 (t-t1)]
a1= 100/ [ ½ 1.82²+ 1.82 (10.4 -1.82)]
a1= 5.79m/s2
We repeat the same calculation for the other Healan runner, whose data are: total distance 100m, acceleration time 3.07 s and total time 10.4 s
Vf2= Vo + a2 t2
Vf2 = x/t
XT = X3 + X4
X3 = Vo t2 + ½ a2 t2²
X3 = ½ a2 t2²
X4 = Vf2 (t-t2)
XT = ½ a2 t2² + Vf2 (t- t2)
XT = ½ a2 t2² + a2 t2 (t-t2)
The aceleration of Healan (a2)
a2 = XT / [½ t2² + t2 (t-t2)]
a2 = 100 / [½ 3,07²+ 3.07 (10.4 -3.07)]
a2 = 3.67 m / s2
We also need the final speeds of each runner
Laura Vf1 = Vo + a1 t1
Vf1 = 0 + 5.79 1.82
Vf1 = 10.54 m / s
Healan Vf2 = Vo + a2 t2
Vf2 = 0 + 3.67 3.07
Vf2 = 11.27 m / s
Having the acceleration and speed of each runner, you can start answering the questions
a) For t3 = 6.15s
Laura
The time to stop with constant speed is what remains after accelerating
XL= ½ a1 t1² + Vf1 (t3-t1)
XL= ½ 5.79 1.82² + 10.54 (6.15 – 1.82)
XL= 55.23 m
Healan
XH= ½ a2 t2² + Vf2 (t3-t2)
XH= ½ 3.67 3.07² + 11.27 (6.15-3.07)
XH= 52.01 m
(XL -XH)= 55.23- 52.01
(XH -XL)= 3.22 m
It is appreciated from these results that Laura is ahead and for a distance of 3.22 m
b) If we analyze the acceleration values of each runner, knowing that they leave the rest and that Healan at the end has a speed greater than Laura, the point of maximum distance difference is when Laura stops accelerating t = 1.82 s
XL= ½ a1 t12
XL= ½ 5.79 1.822
XL= 9.59 m
XH = ½ a2 t12
XH= ½ 3.67 1.822
XH= 6.08 m
The maximum distance difference is 3.51 m
c) Already analyzed in the previous part 1.82 s, since the Laura stop accelerating and Heala continue with acceleration will travel greater distances in equal time units
and the frequency f:
