a car accelerates from rest at a constant rate 'a' for some time. after which it decelerates at a constant rate'b' and comes to rest. if the total time elapsed is t second, evaluate
1) the maximum velocity reached and
2) the total distance travelled.
max velocity=at
distance= 1/2 at^2+at-1/2 bt^2
32 metre
To solve this problem, we can break it down into two parts: the time it takes to accelerate and the time it takes to decelerate.
Let's first consider the acceleration phase. We'll denote the time it takes to accelerate as 't₁'. During this time, the car starts from rest and accelerates at a constant rate 'a'. To find the maximum velocity reached during this phase, we can use the equation of motion:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Since the car starts from rest, the initial velocity (u) is 0. Therefore, the equation simplifies to:
v₁ = a * t₁ ----(1)
Next, we'll consider the deceleration phase. We'll denote the time it takes to decelerate as 't₂'. During this time, the car decelerates at a constant rate 'b'. Again, we can use the equation of motion to find the maximum velocity in this phase:
0 = v₁ - b * t₂ ----(2)
This equation states that the final velocity in the deceleration phase is 0, as the car comes to rest. Rearranging the equation, we get:
v₁ = b * t₂ ----(3)
Since we are given that the total time elapsed is 't' seconds, we can relate the acceleration time (t₁) and deceleration time (t₂) by:
t = t₁ + t₂ ----(4)
To find the maximum velocity reached, we can substitute equations (1) and (3) into equation (2):
0 = (a * t₁) - (b * t₂)
Rearranging the equation, we get:
t₁ = (b / a) * t₂
Substituting this value into equation (4), we get:
t = (b / a) * t₂ + t₂
Simplifying, we get:
t = ((a + b) / a) * t₂
Solving for t₂, we can express it in terms of t:
t₂ = (a / (a + b)) * t
Now that we have the value of t₂, we can substitute it back into equation (3) to find the maximum velocity (v₁) during the deceleration phase:
v₁ = b * [(a / (a + b)) * t] ----(5)
Finally, to find the total distance traveled, we need to sum the distances covered during the acceleration and deceleration phases. The distance covered during each phase can be calculated using the equation:
s = ut + (1/2) * a * t²
For the acceleration phase, the initial velocity (u) is 0 and the acceleration is 'a', so the distance covered (s₁) is:
s₁ = (1/2) * a * t₁² ----(6)
For the deceleration phase, the initial velocity (u) is v₁, the velocity found in equation (5), and the acceleration is 'b', so the distance covered (s₂) is:
s₂ = v₁ * t₂ + (1/2) * (-b) * t₂² ----(7)
Substituting the values of t₁ and t₂ derived earlier into equations (6) and (7) respectively, we can find the distances covered during each phase. The total distance traveled (s) is then the sum of s₁ and s₂:
s = s₁ + s₂
Now you have the complete equations to evaluate the maximum velocity reached (v₁) and the total distance traveled (s) in terms of the given variables 'a', 'b', and 't'.