A car traveling at a constant velocity of 30.0 m/sec passes an intersection at time t=0. 45 seconds later a second car passes the same intersection and its traveling in the same direction at a constant velocity of 55 m/sec.
when does the second car overtake the first car?
how far have the cars traveled when this happens.?
A race car travels northward on a straight, level track at a constant speed travels 0.760 km in 20.0 s. The return trip over the same track is made in 25.0 s.
(a) What is the average velocity of the car in m/s for the first leg of the run?
m/s
(b) What is the average velocity for the total trip?
To determine when the second car overtakes the first car, we first need to find the time it takes for the second car to catch up with the first car.
Let's assume that the second car catches up to the first car at time t. At time t=0, the first car has already traveled a distance of 30.0 m/sec * 0.45 sec = 13.5 meters.
For both cars, distance traveled = velocity * time.
So, the distance traveled by the second car is 55 m/sec * t sec.
Since the second car catches up to the first car at time t and the first car has already traveled 13.5 meters, the distance traveled by the second car when it catches up is also 13.5 meters.
To find when the second car overtakes the first car, we need to solve the equation:
Distance traveled by the second car = Distance traveled by the first car + Distance traveled by the second car when it catches up.
55 m/sec * t sec = 13.5 meters + 55 m/sec * t sec.
Simplifying the equation:
55 m/sec * t sec - 55 m/sec * t sec = 13.5 meters.
0 = 13.5 meters.
However, this equation doesn't provide a solution. This means that the second car will never overtake the first car.
Therefore, the second car will not overtake the first car, and they will remain at the same distance apart.