A car is traveling at a constant speed of 25.5 m/s on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two cars meet for the first time at the next exit, which is 2.35 km away?
To find the acceleration needed for the second car to meet the first car at the next exit, we can use the following steps:
Step 1: Convert the distance to meters
The distance to the next exit is given as 2.35 km. To convert kilometers to meters, we multiply by 1000:
Distance = 2.35 km * 1000 m/km
Distance = 2350 m
Step 2: Determine the time it takes for the first car to reach the next exit
We know that the first car is traveling at a constant speed of 25.5 m/s. We can use the formula: distance = speed * time to determine the time it takes for the first car to reach the next exit.
Time = Distance / Speed
Time = 2350 m / 25.5 m/s
Time = 92.1569 seconds
Step 3: Calculate the final velocity of the second car
The final velocity of the second car is the same as the velocity of the first car since they meet at the next exit.
Final velocity = 25.5 m/s
Step 4: Use the equation of motion to determine the acceleration of the second car
We can use the equation of motion: final velocity = initial velocity + acceleration * time for the second car.
Since the initial velocity is 0 (starting from rest), we can simplify the equation to:
Final velocity = acceleration * time
Acceleration = Final velocity / Time
Acceleration = 25.5 m/s / 92.1569 s
Acceleration ≈ 0.2774 m/s²
Therefore, the second car must maintain an acceleration of approximately 0.2774 m/s² to meet the first car at the next exit.
To find the acceleration that the second car must maintain in order to meet the first car at the next exit, we can use the kinematic equation:
s = ut + (1/2)at^2
Where:
s = distance traveled
u = initial velocity
a = acceleration
t = time taken
For the first car:
s1 = 2.35 km = 2350 m (since 1 km = 1000 m)
u1 = 25.5 m/s
t1 = unknown
For the second car:
s2 = 2350 m
u2 = 0 m/s (started from rest)
a2 = unknown
t2 = unknown
Since both cars meet at the same time at the next exit, their time taken would be the same. Therefore, t1 = t2 = t.
For the first car:
s1 = u1t + (1/2)a1t^2
2350 = 25.5t + (1/2)a1t^2
For the second car:
s2 = u2t + (1/2)a2t^2
2350 = (1/2)a2t^2
Now, we have two equations and two unknowns (a1 and a2). To eliminate the variable t, we can solve one equation for t and substitute it into the other equation.
From the second equation:
t^2 = (4700 / a2)
Substituting this value of t^2 into the first equation:
2350 = 25.5t + (1/2)a1t^2
2350 = 25.5t + (1/2)a1 * (4700 / a2)
2350 = 25.5t + 2350 * (a1 / a2)
Now we can solve for a1:
25.5t = 2350 * (a1 / a2)
a1 = (25.5t * a2) / 2350
So, to find the acceleration that the second car must maintain in order to meet the first car at the next exit, we need the value of t. However, the given information does not provide enough details to calculate the exact value of t. Additional information, such as the time it takes for the first car to travel the 2.35 km, would be required to find the acceleration of the second car.
d = V*t = 2350 m.
25.5*t = 2350.
t = 92.2 s. To reach the next exit.
d = 0.5a*t^2 = 2350 m.
0.5a*92.2^2 = 2350.
4250.42a = 2350.
a = 0.553 m/s^2.