A car starts from rest and accelerates uniformly for 250 meters. It then moves with a constant velocity for 500 meters and slows down to rest in 300 meters. The whole trip takes 50 seconds. How long did the car move at constant velocity?
Not enough info about the acceleration and deceleration. There are an infinite number of solutions
To find out how long the car moved at a constant velocity, we need to understand the motion of the car during each phase and calculate the time taken for each phase separately.
Let's break down the motion of the car into three phases:
1. Acceleration phase: The car starts from rest and accelerates uniformly for 250 meters.
2. Constant velocity phase: The car moves with a constant velocity for 500 meters.
3. Deceleration phase: The car slows down to rest in 300 meters.
First, let's calculate the time taken for the acceleration phase:
We know that the car starts from rest, so initial velocity (u) = 0.
We also know that the car accelerates uniformly, so acceleration (a) remains constant throughout this phase.
The final velocity (v) at the end of this phase is not given, so let's assume it as v1.
Using the formula:
v = u + at
Since the car starts from rest (u = 0), the formula simplifies to:
v1 = at
Given that the car traveled 250 meters during this phase, we can use the formula for distance (s) during uniform acceleration:
s = ut + (1/2)at^2
Since the initial velocity (u) is 0, the formula further simplifies to:
s = (1/2)at^2
Substituting the given values:
250 = (1/2)at^2
Now we can solve this equation for time (t) using the given information.
Once we find the value of t, we can proceed to the next phase.
Next, let's calculate the time taken for the deceleration phase:
Similar to the acceleration phase, we can use the formula for distance during uniform deceleration:
s = ut + (1/2)at^2
Since the car started from a constant velocity during the previous phase, the initial velocity (u) for this phase is v1.
The final velocity (v) is 0, as the car slows down to rest.
Substituting the given values:
300 = v1t + (1/2)(-a)t^2
We have two equations: one from the acceleration phase and another from the deceleration phase. We know the total time taken for both phases is 50 seconds.
So, we can write the equation as:
t1 + t2 = 50
Now we need to substitute the values we have derived in the equations and solve for t1 and t2. Once we find the values of t1 and t2, we can calculate the time taken for the constant velocity phase.
Now, let's calculate the time taken for the constant velocity phase:
The total distance covered during this phase is 500 meters. We can subtract the distances covered during the acceleration and deceleration phases from the total distance travelled:
Constant velocity distance = Total distance - Acceleration distance - Deceleration distance
= 500 - 250 - 300
= 500 - 550
= -50 meters
Since the distance is negative, it means the car did not travel in this phase. Therefore, the time taken for the constant velocity phase is 0 seconds.
In summary, the car did not move at a constant velocity during this trip.