9)You are taking a road trip to the beach for Spring Break. You leave at 12:00pm. At 12:10 you stop for lunch 10 miles down the road. You get back on the road and travel 75 more miles and realize that you need to stop for gas. You notice that it is now 1:10pm. At this point in the trip, what is your average velocity in miles per hour?
11)A race car travels on a racetrack at 38 m/s and slows at a constant rate to a velocity of 20 m/s over 10 s. How far does it move during this time?
d=Vi(t)+1/2(a)(t^2)
d=38(10)+(1/2)(1.8)(100)
I got 470 meters but my choices are:200,290,380,and 760 so I chose 380
On 9) I will be happy to critique your thinking.
on 11) distance while slowing?
average velocity=29
distance=avgvelocity*time....
On your work, a should be negative....
9)I multiplied total distance 85 miles times total time of 70 min. and got 1.21 mi/hr
I divided of course
To calculate the average velocity, you need to find the total distance traveled and the total time taken.
For question 9, let's break down the information given:
- You leave at 12:00 pm.
- At 12:10 pm, you stop for lunch 10 miles down the road.
- Then you travel an additional 75 miles and notice that it is now 1:10 pm.
To find the total distance traveled, you sum up both distances:
- Distance traveled before lunch = 10 miles
- Distance traveled after lunch = 75 miles
- Total distance traveled = 10 miles + 75 miles = 85 miles
To find the total time taken, you subtract the initial time (12:00 pm) from the final time (1:10 pm):
- Time taken = 1:10 pm - 12:00 pm = 1 hour and 10 minutes
However, it's essential to convert the time into hours (since we want to find the velocity in miles per hour). To convert minutes into hours, divide the number of minutes by 60:
- Time taken = 1 hour + 10 minutes ÷ 60 = 1 hour + 0.17 hours = 1.17 hours
Now, calculate the average velocity by dividing the total distance traveled by the total time taken:
- Average velocity = Total distance traveled ÷ Total time taken = 85 miles ÷ 1.17 hours ≈ 72.65 mph
So, your average velocity at that point of the trip is approximately 72.65 miles per hour.
Moving on to question 11, you correctly used the formula for calculating distance:
- d = Vi(t) + 1/2(a)(t^2)
However, it seems there may have been an error in the calculation based on the choices provided. Let's double-check the calculations:
Given:
- Initial velocity (Vi) = 38 m/s
- Final velocity (Vf) = 20 m/s
- Time (t) = 10 s
Using the formula, plug in these values:
- d = Vi(t) + 1/2(a)(t^2)
- d = 38(10) + 1/2(20-38)(10)
- d = 380 + 1/2(-18)(10)
- d = 380 + (1/2)(-180)
- d = 380 - 90
- d = 290 meters
So, the correct distance the race car moves during this time is 290 meters, not 470 meters.
Therefore, for question 11, the correct answer choice is 290 meters, not 380 meters.