Calvin's Driving Trip

We write a system of equations, where $d$ is the length of Lakeshore Drive, and $t$ is the number of hours it took to drive down Lakeshore drive. $d=35t$ $d+4=50(t-\dfrac{1}{10})$ Then, $d=50t-9=35t$ $t=\dfrac{3}{5}$ Finally, $d=35t=\boxed{21}$ .

Let distance covered through Lakeshore Drive be s(miles) , time taken be t(hrs) and given averaged speed is 35mph.

s = 35t ...............(i)

Therefore,

Distance covered through I-94 highway is s + 4 , time taken is t - 0.1 and given averaged speed is 50mph.

(s+4) = 50(t - 0.1)

t = (s+9)/50.

Now, putting value of t in eqn. (i), 35(s+9) = 50s

  7s + 63 = 10s

  3s = 63

    s = 21 (Ans.)

Note that this becomes, by equating time: $\frac{l}{35}=\frac{l+4}{50}+\frac{6}{60}$ Solving for $l$ gives us: $l=21\text{ mi.}$

paragraph 1 x = 35t
paragraph 2 =>(6 min -> hour = 6/60 = 1/10 ) => 4+x = 50(t-1/10) t = 3/5 paragraph 3 x = 35*3/5 = 21

Let the length of the Lakeshore Drive be d and let the time spent to cover that length be t.

In the instance ${d}={35}\times{t}.....(1)$

In the second instance his $distance={d+4}$

and $time={t-0.1}$

Note that the time given in minutes is converted to hours hence 0.1 hours

Thus ${d+4}=50(t-0.1)$

which simplifies to ${50t-d}={9}.....(2)$

from.....(1) ${d}={35t}$ and substituting this into .....(2) we get

${50t}-{35t}={9}$

${15t}={9}$

${t}=\frac{3}{5}$

but ${d}={35t}$

i.e. ${d}={35(\frac{3}{5})}$

Hence the length of Lakeshore Drive is $\boxed{21miles}$

We know that $s=v.t$ where $r$ is length, $v$ is speed, and $t$ is time.

We just have to input the data.

from Lakeshore Drive we got $s=35mil.(tmin)$

from I-94 highway we got $s+4=50mil.(tmin-6min)$

from both equality we put it into one, $s=35mil.(tmin)=50mil.(tmin-6min)$

we got $t=36$

put $t$ to $s=35mil.(tmin)$ or $s+4=50mil.(tmin-6min)$ . And we found $s=21$

Remember that $s$ here is length of Lakeshore Drive and I-94 highway length is $s +4$ .

$v_1 = \frac{s_1}{t_1}$

$v_2 = \frac{s_2}{t_2}=\frac{s_1+\Delta s}{t_1-\Delta t}$

$t_1 = \frac{s_1}{v_1} \implies v_2 =\frac{s_1+\Delta s}{\frac{s_1}{v_1}-\Delta t}$

$v_2 \frac{s_1}{v_1} - \Delta s v_2 = s_1 + \Delta t$

$s_1 \left( \frac{v_2}{v_1} - 1 \right) = \Delta s v_2 + \Delta t$

$s_1 = \frac{\Delta s v_2+ \Delta t}{\frac{v_2}{v_1}-1} = \boxed{21}$

Let's analyze the 2 way journey simultaneously. In the time Calvin completed his return journey, he still had to drive 6 minutes to complete the Lakeshore drive, i.e. 35/10=3.5 miles. But as I-94 highway is 4 miles longer than Lakeshore, the distance between the relative positions of car in both tracks would be 3.5+4=7.5 miles. But the relative velocity of the cars is 50-35=15 miles per hour. If we consider the fact that both cars were at same relative position, i.e. start of each track before the respective journey, we see that the distance between the relative positions of cars increased from 0 to 7.5 when Calvin completed his return drive. As the relative velocity is 15 miles per hour, it took 30 mins for the distance between relative positions of cars to be 7.5 miles. Thus it took Calvin 30 mins to complete the return drive, so length of I-94 highway is 50/2=25 miles. Since Lakeshore is 4 miles shorter, its length is 25-4=21 miles.

distance = speed*time

35t+4 = 50(t-.1) t = .6 hrs

35(.6) = 21 miles

Let the distance traveled by $d$ in time $t$

Therefore, $35 = \dfrac{d}{t}$ or $d = 35t$

By the second condition,

$50 = \dfrac{d+4}{\left( t - \dfrac{6}{60}\right)}$

Thus,

$50t - 5 = d + 4$ or $50t - 5 = 35t + 4$ and $t = \dfrac{9}{15}$

But $d = 35t$ and hence $d = 35 \times \dfrac{9}{15} = \boxed{21}$

let x be the distance and t1 be the time for going to Chicago. Therefore t1=x/35. similarly let t2 be the time of return. Therefore t2=x+4/50. Now according to the condition given, t1-t2=6/60 hrs. On solving for x after substituting the values of t1 and t2, we get the desired answer.