Question

Carl is eating lunch at his favorite cafe when his friend Isaac calls and says he wants to meet him. Isaac is calling from a city 155 miles155 miles away, and wants to meet Carl somewhere between the two locations. Isaac says he will start driving right away, but Carl needs 31.0 min31.0 min to finish his lunch before he can begin driving. Isaac plans to drive at 70.0 mph70.0 mph , and Carl plans to drive at 50.0 mph50.0 mph . Ignore acceleration and assume the highway forms a straight line. How long will Isaac be driving before he meets Carl?

Answer 1

Issac will drive for $\boxed{1.507\text{ hr}}$ before he meets carl.

Further Explanation:

The acceleration of the driving is ignored. It means that the motion of the cars is purely linear on the highways.

Given:

The distance between Carl and Issac is $155\text{ miles}$.

The speed of Carl is $50\text{ mph}$.

The speed of Issac is $70\text{ mph}$.

The time taken by Carl to finish lunch before starting to drive is $31\text{ min}$.

Concept:

Since Carl is eating lunch and he will start after 31 minutes, Issac will cover some distance within this time.

The distance covered by Issac before Carl starts to drive is,

$\begin{aligned}d_1&=\text{speed}\times\text{time}\\&=50\text{ mph}\times\left(\dfrac{31}{60}\right)\text{hr}\\&=36.16\text{ mile}\end{aligned}$

Now after Issac has covered $36.16\text{ miles}$, Carl starts to drive. So, the distance left between Carl and Issac to be covered after carl starts to drive is,

$\begin{aligned}d&=155\text{ miles}-36.16\text{ miles}\\&=118.84\text{ miles}\end{aligned}$

So, now the sum of the distances covered by Issac and Carl in time $t$ should be equal to the $118.84\text{ miles}$.

$\begin{aligned}D_{\text{Issac }}+D_{\text{Carl}}&=118.84\text{ miles}\\(V_{Issac}\times t)+(V_{Carl}\times t)&=118.84\text{ miles}\end{aligned}$

Substituting the values of speed of Issac and Carl,

$\begin{aligned}(70\times t)+(50\times t)&=118.84\text{ miles}\\120t&=118.84\text{ miles}\\t&=\frac{118.84}{120}\text{ hr}\\t&=0.990\text{ hr}\end{aligned}$

So, Issac and Carl meet after $0.990\text{ hr}$ after Carl starts driving.

So, the time for which Issac has been driving is,

$\begin{aligned}T_{\text{ Issac}}&=0.990\text{ hr}+\dfrac{31}{60}\text{ hr}\\&=(0.990+0.516)\text{ hr}\\&=1.507\text{ hr}\end{aligned}$

Thus, Issac will drive for $\boxed{1.507\text{ hr}}$ before he meets carl.

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Answer Details:

Grade: College

Subject: Physics

Chapter: Linear motion in one dimension

Keywords:

Carl, Issac, Speed, meet, 155 miles, eating lunch, 70 mph, 50 mph, 31 min, between two locations, highway forms a straight line.

Answer 2

Answer:

1.507 h

Explanation:

We can think that they move at constant speed, so their positions will follow

X(t) = Xo + v * t

However Carl will start moving 31 minutes later, so we can adjust the equation like this

X(t) = Xo + v * (t - t1)

Where t1 is the time at which each of them will start moving, 0 for Isaac since he starts driving right away, and 31 minutes (0.5167 h) for Carl

Also, we can considre the initial position of Carl to be 0 and his speed to be 50, while Isaac will have an initial position of 155 and a speed of -70 (negative because he is driving towards the origin of coordinates).

Then we have these two equations:

x(t) = 0 + 50 * (t - 0.5167)

x(t) = 155 - 70 * t

These are equations of lines, the point where they intersect determines the place and time they meet.

50 * (t -0.5167) = 155 - 70 * t

50 * t - 25.83 = 155 - 70 * t

120 * t = 180.83

t = 180.83 / 120 = 1.507 h

This is the time Isaac will be driving