a car accelerates from rest at a constant rate x for some time after which it decelaretes at a constant rate y to come to rest .if the total time elapsed is t the total distance travlled by the car is given by what ?
After t1 seconds,
v = x*t1
s = x/2 t1^2
after t-t1 more seconds,
s = (x*t1)(t-t1) - y/2 (t-t1)^2
To find the total distance traveled by the car, we can break down the motion into two parts: the acceleration phase and the deceleration phase.
First, let's consider the acceleration phase:
The car starts from rest, so its initial velocity (u) is 0. The car accelerates at a constant rate (x) for some time (t1).
Using the equation of motion:
distance (s1) = initial velocity (u) * time (t1) + 0.5 * acceleration (x) * time squared (t1)^2
s1 = 0 * t1 + 0.5 * x * t1^2
s1 = 0.5 * x * t1^2
Next, let's consider the deceleration phase:
The car decelerates at a constant rate (y) until it comes to rest. The total time elapsed is t, so the time spent decelerating (t2) is given by t2 = t - t1.
Using the same equation of motion:
distance (s2) = initial velocity (u) * time (t2) + 0.5 * acceleration (y) * time squared (t2)^2
Since the car starts from rest, the initial velocity is 0:
s2 = 0 * t2 + 0.5 * y * t2^2
s2 = 0.5 * y * t2^2
Finally, to find the total distance traveled by the car, we sum up the distances in both phases:
total distance (s) = s1 + s2
s = 0.5 * x * t1^2 + 0.5 * y * t2^2
Therefore, the total distance traveled by the car is given by the equation:
s = 0.5 * x * t1^2 + 0.5 * y * (t - t1)^2