To save fuel, some truck drivers try to maintain a constant speed when possible. A truck traveling at 110.0 km/hr approaches a car stopped at the red light. When the truck is 211.7 meters from the car the light turns green and the car immediately begins to accelerate at 2.30 m/s2 to a final speed of 119.0 km/hr. How close does the truck come to the car assuming the truck does not slow down?
110 km/hr(1000/3600) = 30.6 m/s
119 km/hr = 33.1 m/s
x position of truck = -211.7+ 30.6 t
v car = 2.3 t until v = 33.1
at t = 14.4 s and x = (1/2)(2.3)(14.4^2)
= 238 m
after that xcar = 238 + 33.1 (t-14.4)
so during car acceleration period:
d = Xcar - Xtruck = .5(2.3)t^2 - 211.7 + 30.6 t
or
d = 1.15 t^2 + 30.6 t - 211.7
does that have a minimum between t = 0 and t = 14.4?
I do not know if you do calculus, if not you will have to find the vertex of that parabola with completing the square
d(d)/dt = 0 at min = 2.3 t +30.6
t = 30.6/2.3 = 13.3 seconds to minimum d
then
d = 1.15(13.3^2) +30.6 t -211.7
= 22.3 meters
To find how close the truck comes to the car, we need to determine the stopping distance of the car and compare it to the initial distance between the two vehicles.
First, let's convert the speeds to meters per second (m/s) for consistency.
The truck's speed is 110.0 km/h, so we convert it to m/s:
110.0 km/hr = (110.0 * 1000) / (60 * 60) m/s
= 30.6 m/s (rounded to one decimal place)
The car's final speed is 119.0 km/h, so we convert it to m/s:
119.0 km/hr = (119.0 * 1000) / (60 * 60) m/s
= 33.1 m/s (rounded to one decimal place)
Next, we find the stopping distance (S) of the car using the equation:
S = (v^2 - u^2) / (2a)
Where v is the final velocity of the car, u is the initial velocity of the car, and a is the acceleration of the car.
Using the given values:
v = 33.1 m/s
u = 0 m/s (the car was initially stopped)
a = 2.30 m/s^2
S = (33.1^2 - 0^2) / (2 * 2.30)
S = (1095.61) / 4.60
S = 238.38 m (rounded to two decimal places)
Now, we can find how close the truck comes to the car by comparing the stopping distance of the car (238.38 m) to the initial distance between them (211.7 m).
The truck approaches the car by 211.7 m, which is less than the stopping distance of the car (238.38 m).
Therefore, the truck would come within 211.7 meters of the car, assuming it does not slow down.
To solve this problem, we can break it down into two parts: determining the distance it takes for the car to reach its final speed and the distance the truck travels during that time.
First, let's find out how long it takes for the car to accelerate from rest (0 km/hr) to its final speed of 119.0 km/hr. We can use the formula:
v = u + at,
where:
- v is the final velocity (119.0 km/hr),
- u is the initial velocity (0 km/hr),
- a is the acceleration (2.30 m/s^2), and
- t is the time.
Converting the velocities to meters per second (m/s), we have:
119.0 km/hr = (119.0 * 1000) m / (60 * 60) s = 33.06 m/s,
0 km/hr = (0 * 1000) m / (60 * 60) s = 0 m/s.
Now we can rearrange the formula to solve for t:
t = (v - u) / a.
Substituting the values, we get:
t = (33.06 - 0) m/s / 2.30 m/s^2 = 14.37 s.
Now we know that it takes the car 14.37 seconds to reach its final speed.
To find out how far the truck travels during this time, we need to calculate the distance covered by the truck. We can use the equation:
distance = speed × time.
Converting the truck's speed of 110.0 km/hr into m/s:
110.0 km/hr = (110.0 * 1000) m / (60 * 60) s = 30.56 m/s.
Multiplying the speed by the time:
distance = 30.56 m/s × 14.37 s = 439.45 m.
Therefore, the truck travels a distance of 439.45 meters during the time it takes for the car to accelerate.
Next, we want to calculate how close the truck comes to the car. We can find this by subtracting the distance traveled by the truck from the initial distance between the truck and the car.
Initial distance = 211.7 m
Final distance = 211.7 m - 439.45 m = -227.75 m.
The negative value indicates that the truck passes the car and ends up 227.75 meters behind it.
So, the truck comes approximately 227.75 meters close to the car assuming it does not slow down.