Can anyone help me figure out how to answer this physics problem?

A cab driver picks up a customer and delivers her 1.47 km away, driving a straight route. The driver accelerates to the speed limit and, upon reaching it, begins to decelerate immediately. The magnitude of the deceleration is 3.11 times the magnitude of the acceleration. Find the lengths of the (a) acceleration and (b) deceleration phases of the trip.Can anyone help me figure out how to answer this physics problem?
Start by breaking the process down into two steps: one with positive acceleration, and another with negative acceleration (deceleration). The total distance traveled is 1.47 km, so we'll consider the distance traveled during acceleration as x, and the distance traveled during decelration as (1.47 - x).





We need to start with the formula for velocity and acceleration:





v = v(0) + a*t





Our v is the speed limit, and can be the same for both acceleration and deceleration. Hopefully I can illustrate this ...





Our v(0) is 0 for acceleration, since the driver starts from a dead stop. The initial acceleration equation reduces to





v = a*t1, where t1 is the time of acceleration.





Deceleration is just negative acceleration. So the time it takes the driver to decelerate from the speed limit to a stop is also the same time it takes the driver to accelerate to the speed limit from a dead stop. The only difference is the rate of deceleration. We know the rate of deceleration is 3.11 time that of acceleration, so our deceleration equation becomes





v = 3.11a*t2, where t2 is the time of decelration.





Now we need to relate t1 and t2:





v = a*t1 = 3.11*a*t2, which means





t1 = 3.11*t2 (that is, it takes 3.11 times as long to accelerate to the speed limit as it does to decelerate from it).





Now, let's consider the equation that relates distance and acceleration:





distance 1 = x = x(0) + v(0)*t+(1/2)*a*(t1)虏.


We know that the driver starts from a stop, so the initial velocity v(0)=0, and we'll consider his starting position x(0) as (0). This reduces the equation to





x = (1/2)*a*(t1)虏





distance 2: as mentioned in the velocity equation, we can express v(0) as 0. Our distance traveled is (1.47 - x), and if we use the point of deceleration as our starting point, then we can express x(0) as 0, and the equation becomes





1.47 - x = (1/2)*3.11*a*(t2)虏





x = 1.47 - [1.555a*(t2)虏]





(remember that our deceleration is 3.11 times that of our acceleration). If we substitute t2 for t1, our equations become





x = (1/2)*a*(3.11*t2)虏


x = 1.47 - [1.555a*(t2)虏],





and we can set them equal to one another





1.47 - [1.555*a*(t2)虏] = 4.83605*a*(t2)虏





6.39105*a*(t2)虏 = 1.47





a*(t2)虏 = 0.23





We can take this value and plug it back into either equation, but the second one looks easiest:





x = 1.47 - [1.555a*(t2)虏]





Solve for x, and you have your acceleration length.


Solve for (1.47 - x) and you have your deceleration length.





I hope this was helpful.