Need help with physics questions!!?

Question

Im very very lost. I have this question to answer "A person takes a trip, driving with a constant speed of 97.5 km/h except for a 20.0 min rest stop. If the person's average speed is 66.5 km/h, how much time is spent on the trip and how far does the person travel?

"

I have to find the total h the person spent traveling, and the total km distance. Well I have this

"I have Stot=(d1+0)/(t1+.333sec)

And S1=d1/t1 which I have as 97.5=x/y

and S2=d2/t2 which I have as 0=0/z

"

And my professor said

"Ignore the 3rd equation which just amounts to 0=0 and doesn't help. Solve the 2nd equation for d1 (i.e. d1=...). Replace d1 in the 1st equation with the d1 equation that you just found. You should now have a single equation with one unknown, t1."

The problem is I totally do not know how to solve for d1 when I dont have t1.

My other question is "The average person passes out at an acceleration of 7g (that is, seven times the gravitational acceleration on Earth). Suppose a car is designed to accelerate at this rate. How much time (in seconds) would be required for the car to accelerate from rest to 58.1 miles per hour? (The car would need rocket boosters!)"

This one I totally do not know how to do -.- Im just not meant for physics

Answer 1

You have written 0.333 sec. As I understand you divided 20 by 60 and got 0.333. That is correct. But it should be 0.333 hour. Why did you write sec? I think it is typo.

Here is the solution:-

In the following, distance is in km and time is in h.

Stot =(d1+0)/(t1+.333)

Or 66.5 = d1/(t1+0.333)----------------(1)

S1 = d1/t1

Or 97.5 = d1/t1

Or d1 = 97.5t1--------------(2)

Substituting the value of d1 from (2) into (1),

66.5 = 97.5t1/(t1+0.333)

Or 66.5*(t1+0.333) = 97.5t1

Or 66.5t1 + 22.14 = 97.5t1

Or 97.5t1 - 66.5t1 = 22.14

Or 31t1 = 22.14

Or t1 = 22.14/31

Or t1 = 0.71 hour

From (2),

d1 = 97.5t1

Or d1 = 97.5 * 0.71 km

= 69.22 km

Answer 2

Ok, lets go through the typical problem solving for a good explanation:

1) List the knowns and unknowns

Constant speed = 97.5 km/h

Rest time = 20 min

Average speed = 66.5 km/h

Uknown= time spent on trip

Unknown =distance traveled

2) List relationships that use the knowns and unknowns in an effort to find a solution

Speed of whole trip = Dist of trip /(rest + time spent driving)

Speed of driving = Dist of trip/time spent driving

66.5 km/hr = Dist of Trip /( 0.333 hr + time spent driving)

97.5 km/hr = Dist of Trip/(time spent driving)

So, I see two equations with the same two unknowns

Solve the 2nd one for Dist of trip, you say you don' tknow how, you just need to do a bit of algebra

97.5 km/hr = Dist of Trip/time spent driving

97.5 * time spent driving = Dist of Trip / time spent driving * time spent driving (just do the same thing to both sides of the equation)

97/5 * time spent driving = Dist of Trip

take this and substitue into the first equation

66.5 km/hr = 97.5 * time spent driving / (0.333 hr + time spent driving)

Now just solve for Time spent Driving and sub that back into the second equation and solve for dist of trip and you are done

The second quation

OK, accelleration is pretty easy, look at the units

meters/second/second for example.  Gravity is 9.8 m/s per second so each second that goes by it accelerates another 9.8 m/s.

Lets use metric unilts .  Again with the knowns and unknowns

Gravity = 9.8 m/s^2

Acceleration rate = 7 g = 7* 9.8 m/s^2

Goal velocity = 58.1 mph = 58.1 mph * 5280 ft/mile * 12 inches/ft * 2.54 cm/in * 1 m/100 cm * 1 hr/60 min * 1 min/60 sec = 25.973 m/s

Now for the equation

Speed = acceleration * time

solve for time

time = speed/acceleration

=25.973/(9.8*7)

=0.379 sec

Thats about 4x as fast as a reaaalllly fast sports car, I think faster than a dragster