Question 1197004: A car and a truck leave the same place and travel in the same direction along a straight road. The car starting from rest speeds up to 24 kph with a constant acceleration of 1/6 * m / (s ^ 2) and then runs at this speed. The truck leaves 40 seconds after the car with a uniform acceleration of 1/3 * m / (s ^ 2) from rest to attain a speed of 48 kph and then travels at this speed. How soon after the car started will the truck overtake the car?
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
t = number of seconds the car has been driving
kph = kilometers per hour
1 km = 1000 meters
1 min = 60 sec
1 hour = 60 min
1 hour = (60*60) sec
1 hour = 3600 sec
Because the acceleration is in m/s^2, but the velocities are in kph, I'll convert 24 kph to m/s
We'll use a kinematics equation to find the time it takes for the car to reach a velocity of (20/3) m/s.
Vi = initial velocity = 0
Vf = final velocity = 20/3
a = acceleration = 1/6
t = (Vf - Vi)/a
t = (20/3 - 0)/(1/6)
t = 40
It takes 40 seconds for the car to reach 20/3 meters per second, aka 24 kph
When the truck starts up, the car has reached its target speed of 24 kph.
Now use another kinematics equation to compute the distance traveled during this time for the car.
d = ( (vf)^2 - (vi)^2 )/(2a)
d = ( (20/3)^2 - (0)^2 )/(2*1/6)
d = 400/3
The car has traveled 400/3 meters over the course of 40 seconds when it was accelerating from 0 to 24 kph.
As the car is accelerating, we cannot use distance = rate*time since the rate aka speed is changing.
After the t = 40 second mark, the car is now traveling at a constant velocity.
At this point we can use distance = rate*time
The car travels a further (20/3)*(t-40) meters where t > 40
In total the car travels a distance of 400/3 + (20/3)*(t-40) meters when t > 40
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Now we move to the truck. We'll follow the same outline as we did with the car in the previous section.
or note that because 48 = 2*24, i.e. the truck's final velocity is twice that of the car's final velocity, we can say: 2*(20/3) = 40/3
Calculate the time duration throughout the acceleration period
Vi = initial velocity = 0
Vf = final velocity = 40/3
a = acceleration = 1/3
t = (Vf - Vi)/a
t = (40/3 - 0)/(1/3)
t = 40
The truck also needs 40 seconds to accelerate to the target velocity it's after.
This is expected because the truck is going twice as fast, and its acceleration is twice as much
2*(1/6) = 1/3
Effectively the '2's cancel out and that's why we end up with t = 40 like from before.
Then calculate the distance during this acceleration period
d = ( (Vf)^2 - (Vi)^2 )/(2a)
d = ( (40/3)^2 - (0)^2 )/(2*1/3)
d = 800/3
The total distance the truck travels is
800/3 + (40/3)*(t-80) where t > 80
The t-80 refers to the idea where the truck waited 40 seconds after the car started.
Then it takes another 40 more seconds for the truck to get to the speed of 40/3 meters per second (48 kph)
In total, the truck has reached the timestamp of 40+40 = 80 seconds once it has reached the desired velocity.
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We have these two distance expressions
Car: 400/3 + (20/3)*(t-40) when t > 40
Truck: 800/3 + (40/3)*(t-80) when t > 80
For both of them to make sense, we need t > 80, which is where the two intervals overlap.
Set the two expressions equal to one another to see when they'll meet up
400/3 + (20/3)*(t-40) = 800/3 + (40/3)*(t-80)
3 * [ 400/3 + (20/3)*(t-40) ] = 3 * [ 800/3 + (40/3)*(t-80) ]
400 + 20*(t-40) = 800 + 40*(t-80)
400 + 20t - 800 = 800 + 40t - 3200
20t - 400 = 40t - 2400
20t-40t = -2400+400
-20t = -2000
t = -2000/(-20)
t = 100
It takes 100 seconds for the truck to meet up with the car. After this point in time the truck passes by the car.
Recall that the start point is from the car's perspective. In other words, once the car starts driving is when the clock starts. The truck will have been driving for t-40 = 100-40 = 60 seconds when the truck meets the car, due to the 40 second head start the car had.
Answer: 100 seconds
Extra info:
100 seconds = 60 seconds + 40 seconds = 1 minute + 40 seconds
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