the escalator that leads down into a subway station has a length of 30.0m and a speed of 1.8 m/s relative to the ground. student is coming out of subway by running up the wrong direction on escalator. local record time for this is 11s. relative to escalator, what speed must the student exceed in order to beat the record?

(speed-1.8)time=distance

solve for speed.

is this correct?:

speed(time)=distance

speed=dist/time

Of course not. I gave you the answer above.

speed-1.8 = distance/time

that not the answer i have in my book. the answer i have is 4.5m/s.

You need to get a basic math tutor immediately, or drop physics. The language of physics is math, and you are operating at an algebra I level, at best. You will not succeed.

speed-1.8=30/11
speed= 30/11 + 1.8= 4.5m/s

Physics comprehension will demand much higher levels of math skills than this.

Good luck.

Hi , I know how you did all the mathematically calculations but I was just wondering how u came up with the eqauation that you did.

To solve this problem, we need to understand the concept of relative velocities and how they impact the movement of objects.

The speed of the escalator is given as 1.8 m/s. Let's call this v_esc, where v_esc = 1.8 m/s. The length of the escalator is given as 30.0 m.

Now, let's consider the movement of the student running up the escalator. The student is essentially moving against the direction of the escalator, so their effective speed will be the difference between their actual speed and the speed of the escalator.

If we denote the speed of the student as v_stu, then the relative speed of the student with respect to the ground (which is his effective speed) can be calculated as:

v_eff = v_stu - v_esc

To beat the local record time for coming out of the subway, the student needs to complete the distance of 30.0 m in less than 11 seconds. We can use the formula for speed:

Speed = Distance / Time

To beat the record, the student needs to exceed this speed. So the inequality becomes:

v_stu - v_esc > Distance / Time

To find the minimum speed that the student needs to exceed, we can plug in the given values:

v_stu - 1.8 > 30.0 / 11

Now, we can solve this inequality to find the minimum speed. First, let's isolate v_stu:

v_stu > 30.0 / 11 + 1.8

v_stu > 2.727 m/s (rounded to three decimal places)

Therefore, the student must exceed a speed of 2.727 m/s relative to the escalator in order to beat the local record time.