a car accelerates from rest at a constant rate alpha for some time, after which it decelerates at a constant rate beeta and comes to rest. if the total time elapsed is t, the maximum velocity acquired by the car will be?
To find the maximum velocity acquired by the car, we need to break down the motion into two parts: the acceleration phase and the deceleration phase.
Let's consider the acceleration phase first. The car starts from rest and accelerates at a constant rate α. We can use the equation for motion with constant acceleration to find the time taken during this phase. Since the initial velocity (u) is zero, the equation becomes:
v = u + at
where v is the velocity, a is the acceleration, and t is the time. Rearranging the equation, we get:
t = (v - u) / a
During the acceleration phase, the final velocity (v) will be the maximum velocity, which we'll call vmax. Thus, the equation becomes:
t = (vmax - 0) / α
t = vmax / α
Now, let's consider the deceleration phase. The car decelerates at a constant rate β until it comes to rest. Using the same equation as before, we find the time taken during this phase:
t = (0 - vfinal) / β
t = -vfinal / β
Since the total time elapsed is t, we can sum the times taken during the two phases:
total time = t + (vmax / α) + (-vfinal / β)
We know that the total time elapsed is t, so we can substitute that value in:
t = t + (vmax / α) + (-vfinal / β)
Now, let's rearrange this equation to isolate vmax:
(vmax / α) + (-vfinal / β) = 0
Multiplying through by αβ, we get:
βvmax - αvfinal = 0
Now, since the final velocity is zero when the car comes to rest, we can substitute that in:
βvmax - α(0) = 0
βvmax = 0
Thus, vmax = 0. Therefore, the maximum velocity acquired by the car is zero.
In simpler terms, if a car first accelerates from rest and then decelerates to come to rest, the maximum velocity it can achieve is zero, indicating that the car doesn't reach any maximum speed.
Let maximum velocity be x
then x/α+x/β=t
Solving,
x=t/(1/α+1/β)
=αβt/(α+β)