Two cars travel westward along a straight highway, one at a constant velocity of 85 km/hr, and the other at a constant velocity of 115 km/hr.
a) Sketch the velocity vs. time graph for both cars and paste your sketch in the space below.
b) Assuming both cars start at the same point, how much sooner does the faster car arrive at a destination 16 km away?
c)How far must the cars travel for the faster car to arrive 15 min before the slower car?
Which car was travelling at steady speed
a) I wish I could draw, but I'm just a bot. However, I can describe it for you! The velocity vs. time graph for both cars would be a straight line, since their velocity is constant. The line for the car traveling at 85 km/hr would have a lower slope than the line for the car traveling at 115 km/hr, since it's slower.
b) To find out how much sooner the faster car arrives at a destination 16 km away, we can divide the distance by the difference in their velocities. In this case, (115 km/hr - 85 km/hr) is equal to 30 km/hr. So, the faster car would arrive 16 km sooner divided by 30 km/hr, which is approximately 0.53 hours, or 32 minutes.
c) To calculate how far the cars must travel for the faster car to arrive 15 minutes before the slower car, we first need to convert the time difference to hours. 15 minutes is equal to 0.25 hours.
Next, we need to calculate the difference in velocity again, which is 115 km/hr - 85 km/hr, which is equal to 30 km/hr.
Then, we can calculate the distance by multiplying the time difference by the velocity difference: 0.25 hours * 30 km/hr, which equals 7.5 km.
So, the faster car must travel 7.5 km further than the slower car for it to arrive 15 minutes before the slower car.
a) To sketch the velocity vs. time graph for both cars, we need to understand that velocity is the rate at which an object changes its position. The graph will show how the velocity of each car changes over time.
For the car traveling at a constant velocity of 85 km/hr, the velocity will be constant throughout. Therefore, the graph will be a straight line parallel to the time axis at a height of 85 km/hr.
For the car traveling at a constant velocity of 115 km/hr, the graph will be another straight line parallel to the time axis but at a height of 115 km/hr.
Sketching these lines on a graph, we can label the vertical axis as velocity (in km/hr) and the horizontal axis as time (in hours).
b) To determine how much sooner the faster car arrives at a destination 16 km away, we can calculate the time it takes for each car to reach that distance.
For the first car traveling at 85 km/hr:
Time = Distance / Velocity = 16 km / 85 km/hr = 0.188 hours
For the second car traveling at 115 km/hr:
Time = Distance / Velocity = 16 km / 115 km/hr = 0.139 hours
The faster car takes 0.139 hours to reach the destination, while the slower car takes 0.188 hours. The time difference is:
Time difference = 0.188 hours - 0.139 hours = 0.049 hours
Therefore, the faster car arrives at the destination 16 km away approximately 0.049 hours (or 2.94 minutes) sooner.
c) To determine the distance the cars must travel for the faster car to arrive 15 minutes (or 0.25 hours) before the slower car, we can set up an equation based on the time difference.
Let "x" be the distance both cars need to travel.
For the first car traveling at 85 km/hr:
Time = Distance / Velocity = x km / 85 km/hr
For the second car traveling at 115 km/hr:
Time = Distance / Velocity = (x - 16) km / 115 km/hr
Since we want the time difference to be 0.25 hours (or 15 minutes), we can set up the equation:
(x - 16) km / 115 km/hr - x km / 85 km/hr = 0.25 hours
To solve this equation, we can simplify it by finding a common denominator for the fractions and then cross-multiplying:
85(x - 16) - 115x = 0.25 * 85 * 115
Simplifying further:
85x - 1360 - 115x = 2473.75
-30x - 1360 = 2473.75
-30x = 3833.75
x ≈ -127.79
Since distance cannot be negative, this result does not make sense in this context. Therefore, it is not possible for the faster car to arrive 15 minutes before the slower car.
b. d1 = 85t1 = 16 km,
85t1 = 16,
t1 = 16 / 85 = 0.1882 h = 11.29 min.
d2 = 115t2 = 16 km,
115t2 = 16,
t2 = 16 / 115 = 0.139 h = 8.35 min.
t1 - t2 = 11.29 - 8.35 = 2.94 min.
Car #2 arrives 2.94 min. sooner.
c. 15 min = 1/4 h = 0.25 h.
d1 = d2,
115t = 85(t + 0.25),
115t = 85t + 21.25,
115t - 85t = 21.25,
30t = 21.25,
t = 21.25 / 30 = 0.71 h,
d = 115 * 0.71 = 81.65 km.