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.
the answer for this problem is
1) abt/a+b
2) abt^2/2(a+b)
pls tell me how this answer is arrived at
Your given answer for (2) can't be correct, it doesn't have units of distance (it has units of acceleration).
In general, v = v_0 + a t, where a is any acceleration.
Let t1 be the time during the acceleration and t2 be the time during the deceleration and v = the speed at the end of the acceleration.
v - 0 = a t1
0 - v = -b t2
v = a t1 = b t2
This yields the equation
a t1 - b t2 = 0
We also know that
t1 + t2 = t
These are two linear equations in two unknowns. There are several ways of solving these. One way is to solve the first equation for t2 in terms of t1, then plug this into the second equation:
t2 = (a/b)t1
t1 + (a/b)t1 = t
(b + a)/b t1 = t
t1 = b/(a + b) t
Similarly, t2 = a/(a + b) t
From one of the first equations
v = a t1 = a[b/(a + b) t] = abt/(a+b)
As I mentioned in an earlier comment, your answer for (2) cannot be correct.
To arrive at the given answers, we'll use the equations of motion for uniformly accelerated motion.
Let's break down the problem into two parts: the acceleration phase and the deceleration phase.
1) Maximum velocity reached:
During the acceleration phase, the car starts from rest and accelerates at a constant rate 'a' for a certain time. We can use the equation of motion:
velocity (v) = initial velocity (u) + acceleration (a) * time (t)
Initially, the velocity is zero (as the car starts from rest), so we have:
v₁ = 0 + a * t₁
During the deceleration phase, the car decelerates at a constant rate 'b' and comes to rest. We can use the same equation of motion, but this time the final velocity is zero.
0 = v₁ + b * t₂
Rearranging this equation, we get:
t₂ = -v₁ / b
Now, let's substitute the value of v₁ from the acceleration phase equation into the deceleration phase equation:
t₂ = -(a * t₁) / b
The total time elapsed (t) is the sum of the time taken during the acceleration and deceleration phases:
t = t₁ + t₂
Substituting the value of t₂, we get:
t = t₁ - (a * t₁) / b
Rearranging this equation, we can find t₁ in terms of t:
t₁ = (bt) / (a + b)
Substituting this value of t₁ back into the acceleration phase equation, we can find the maximum velocity (v max):
v max = a * t₁ = a * (bt) / (a + b) = abt / (a + b)
So, the maximum velocity reached is abt / (a + b).
2) Total distance travelled:
The distance traveled during uniformly accelerated motion can be calculated using the equation:
distance (s) = initial velocity (u) * time (t) + 1/2 * acceleration (a) * time (t)^2
During the acceleration phase, the initial velocity (u₁) is zero, so we have:
s₁ = 0 + 1/2 * a * t₁^2 = 1/2 * a * (bt / (a + b))^2 = (abt^2) / (2(a + b)^2)
During the deceleration phase, the final velocity is zero, so the distance traveled during this phase is zero.
Therefore, the total distance traveled (s total) is the sum of the distances traveled during the acceleration and deceleration phases:
s total = s₁ + s₂ = (abt^2) / (2(a + b)^2) + 0 = (abt^2) / (2(a + b))
Hence, the total distance traveled is (abt^2) / (2(a + b)).