Two cars are driving at the same constant speed on a straight road, with car 1 in front of car 2. Car 1 suddenly starts to brake with constant acceleration and stops in 10 m. At the instant car 1 comes to a stop, car 2 begins to brake with the same accelaration. It comes to a halt just as it reaches the back of car 1. Whar was the separation between the cars before they started braking?
Let's denote the initial separation between the cars as "d" and the constant acceleration of both cars as "a".
Car 1 comes to a stop in 10 m with constant acceleration, which means we can use the following equation of motion:
Vf^2 = Vi^2 + 2ad
Since car 1 comes to a stop, its final velocity (Vf) is 0. We can assume that its initial velocity (Vi) is also unknown. However, we can rewrite the equation as:
0 = Vi^2 + 2ad
Similarly, when car 2 comes to a stop just as it reaches the back of car 1, the final separation between the cars is 0. We can use the same equation of motion to find the initial velocity (Vi2) of car 2:
0 = Vi2^2 + 2ad
Since both cars have the same constant acceleration (a), we can equate the two equations:
Vi^2 + 2ad = Vi2^2 + 2ad
Simplifying the equation, we find:
Vi^2 = Vi2^2
This implies that the initial velocities of both cars are equal. Let's denote this common initial velocity as "V".
Now, let's consider the relative motion of the two cars. When car 2 starts braking, it has the same initial velocity (V) as car 1 does when it starts braking. The relative velocity, denoted as V_rel, between the two cars is:
V_rel = V - (-V) = 2V
Now, we can find the time (t) it takes for car 2 to catch up to car 1. Using the equation of motion:
d = V_rel * t + (1/2) * a * t^2
Since both cars have the same acceleration and car 2 stops at the same time as it reaches the back of car 1, we can substitute "a" and "d" with their respective values:
0 = 2V * t + (1/2) * a * t^2
Simplifying the equation, we find:
a * t^2 = -4V * t
Dividing both sides by t and rearranging, we find:
at = -4V
Now, we can use the equation of motion for car 1 to find the time it takes for car 1 to stop:
0 = V^2 + 2ad
Simplifying, we find:
ad = -V^2
Substituting "-V^2" with "ad", we have:
a * d = -V^2
Now, we can rearrange the equation and substitute "-4V" for "at":
d = (-V^2) / a = -V * (-4V) / a = 4V^2 / a
Since we've previously determined that the initial velocities of both cars are equal, we can substitute "V" with "Vi" or "Vi2". Therefore, the separation between the cars before they started braking is:
d = 4Vi^2 / a
Therefore, the separation between the cars before they started braking is given by the equation d = 4Vi^2 / a.
To find the separation between the cars before they started braking, we can set up equations of motion for both cars and solve for the unknown variables.
Let:
- s1 = initial separation between car 1 and car 2 (what we want to find)
- a = acceleration (which is the same for both cars as mentioned)
- t1 = time taken by car 1 to come to a stop
- t2 = time taken by car 2 to come to a stop
For Car 1:
Using the equation of motion s = u*t + (1/2)*a*t^2, where u is the initial velocity (which we assume is constant),
we have:
0 = u*t1 + (1/2)*a*t1^2 -- Equation 1
For Car 2:
The initial velocity of car 2 is the final velocity of car 1 when it comes to a stop. Since both cars have the same constant acceleration, the final velocity of car 1 is 0. Therefore, we have:
u = a*t1 -- Equation 2
Using Equation 2, we can express u in terms of a and t1:
u = a*t1 -- Substitute into Equation 1
0 = a*t1^2 + (1/2)*a*t1^2
0 = (3/2)*a*t1^2
Since the acceleration "a" cannot be zero (otherwise the cars will not move), we have:
t1^2 = 0 -- which implies t1 = 0
However, this is not a valid solution since time cannot be zero in this context. So, we conclude that Equation 2 does not hold in this scenario.
Let's consider a different approach.
Let v1 be the velocity of car 1 when it comes to a stop.
v1 = u + a*t1 -- Equation 3
For Car 2:
Using the equation of motion s = u*t + (1/2)*a*t^2:
s1 = v1*t2 + (1/2)*a*t2^2 -- Equation 4
We can rewrite Equation 4 using Equation 3:
s1 = (u + a*t1)*t2 + (1/2)*a*t2^2
Since u = a*t1 (from Equation 2), we have:
s1 = (a*t1 + a*t1)*t2 + (1/2)*a*t2^2
s1 = 2*a*t1*t2 + (1/2)*a*t2^2
s1 = a*t2*(2*t1 + (1/2)*t2) -- Equation 5
From Equation 1, we know that:
0 = u*t1 + (1/2)*a*t1^2
u*t1 = -(1/2)*a*t1^2 -- Equation 6
Using Equation 6, we can express u in terms of a and t1:
u = -(1/2)*a*t1 -- Substitute into Equation 5
s1 = a*t2*(2*t1 + (1/2)*t2)
We know that s1 = 10 m, and we want to find t2 in terms of t1. We also need to consider that t1 and t2 cannot be zero since the cars are in motion.
To proceed further, we need more information, such as the values of t1 or t2, or the value of the acceleration "a". Without this additional information, we can't determine the exact separation between the cars before they started braking.
the average speed during braking is half of the constant driving speed
so car 2 travels 20 m in the time it takes car 1 10 m to stop
... this moves car 2 10 m closer to car 1
car 2 then travels an additional 10 m while stopping