"A cab driver picks up a customer and delivers her 3.31 km away, driving a straight route. The driver accelerates to the speed limit and, upon reaching it, begins to decelerate immediately. The magnitude of the deceleration is 4.70 times the magnitude of the acceleration. Find the lengths of the (a) acceleration and (b) deceleration phases of the trip."
To find the lengths of the acceleration and deceleration phases of the trip, we can use the equations of motion.
Let's define the following variables:
- s = distance covered
- u = initial velocity
- v = final velocity
- a = acceleration
- d = deceleration
- t = time
First, we need to find the initial velocity, final velocity, and time of the trip.
Given that the driver accelerates to the speed limit and then begins decelerating immediately, we can assume that the final velocity is zero (v = 0). We also know that the distance covered is 3.31 km (s = 3.31 km).
Using the equation of motion:
v^2 = u^2 + 2as
And assuming the initial velocity (u) is zero since the driver starts from rest, we can rewrite the equation as:
v^2 = 2as
Plugging in the given values:
0^2 = 2a(3.31)
Simplifying further:
0 = 6.62a
Since the final velocity is zero, the speed limit is reached when the driver has covered half the distance of 3.31 km, which is 1.655 km.
Now let's calculate the acceleration phase:
Using the equation:
v = u + at
Since u = 0 and v = 0, we get:
0 = 0 + a(t1)
Since the final velocity is zero, the time taken during the acceleration phase (t1) is zero.
Next, let's calculate the deceleration phase:
Using the equation:
v = u + at
Since u = 0 and v = 0, we get:
0 = 0 + d(t2)
Since the final velocity is zero, the time taken during the deceleration phase (t2) is zero.
Therefore, the lengths of both the acceleration and deceleration phases are zero.
In conclusion:
(a) The length of the acceleration phase is 0 km.
(b) The length of the deceleration phase is 0 km.
To find the lengths of the acceleration and deceleration phases of the trip, we can use the following steps:
Step 1: Convert the distance traveled into meters:
3.31 km = 3310 meters
Step 2: Let's assume that the time taken for both acceleration and deceleration phases is the same.
Let's say this time taken is 't'.
Step 3: Calculate the acceleration 'a' during the acceleration phase:
Using the third equation of motion:
s = ut + (1/2)at^2
Here, s = 3310 meters (distance), u = 0 m/s (initial velocity), t = t (time), and a = acceleration.
Since the initial velocity is 0 m/s and the final velocity is the speed limit, we can rewrite the equation as:
3310 = 0t + (1/2)a*t^2
Simplifying further, we get:
3310 = (1/2)a*t^2 ------- (Equation 1)
Step 4: Calculate the deceleration 'd' during the deceleration phase:
Given that the magnitude of the deceleration is 4.70 times the magnitude of the acceleration, we can write:
d = 4.70a
Step 5: Calculate the distance covered during the acceleration phase:
Using the first equation of motion:
v = u + at
Since the final velocity during the acceleration phase is the speed limit and the initial velocity is 0, we have:
v = a*t
The distance covered during the acceleration phase (d_acc) can be given as:
d_acc = (1/2)(a*t)
Step 6: Calculate the distance covered during the deceleration phase:
The distance covered during the deceleration phase (d_dec) can be given as:
d_dec = (1/2)(d*t)
Step 7: Equate the total distance with the sum of the distances covered during the acceleration and deceleration phases:
3310 = d_acc + d_dec
Substituting the values obtained from steps 5 and 6 into the equation above, we get:
3310 = (1/2)(a*t) + (1/2)(d*t)
Step 8: Solve the equation to find the value of 't':
3310 = (1/2)(a + d) * t
Dividing both sides of the equation by (1/2)(a + d), we get:
t = 2 * (3310 / (a + d)) ------- (Equation 2)
Step 9: Substitute the value of 't' from Equation 2 into Equation 1 and solve for 'a':
3310 = (1/2)a(t^2)
3310 = (1/2)a * (2*(3310 / (a + d)))^2
Simplifying further, we get:
3310 = a * (4*(3310^2) / (a + d)^2)
Step 10: Solve the equation obtained in Step 9 to find 'a'.
Multiply both sides of the equation by (a + d)^2 and simplify:
3310 * (a + d)^2 = 4 * 3310^2 * a
Expand and simplify:
3310 * (a^2 + 2ad + d^2) = 4 * 3310^2 * a
3310 * a^2 + 6620 * ad + 3310 * d^2 = 4 * 3310^2 * a
3310 * a^2 - 4 * 3310^2 * a + 6620 * ad + 3310 * d^2 = 0
Next, we can substitute d = 4.70a into the equation above:
3310 * a^2 - 4 * 3310^2 * a + 6620 * a * (4.70a) + 3310 * (4.70a)^2 = 0
Simplify the equation to:
3310 * a^2 - 4 * 3310^2 * a + 6620 * 4.70 * a^2 + 3310 * 4.70^2 * a^2 = 0
Combine like terms:
3310 * a^2 - (4 * 3310^2 - 6620 * 4.70 - 3310 * 4.70^2) * a^2 = 0
Simplify further:
3310 * a^2 - (4.70^2 - 2 * 4 * 4.70 + 1) * 3310^2 * a^2 = 0
This equation can be simplified to:
3310 * a^2 - (4.70^2 - 8 * 4.70 + 1) * 3310^2 * a^2 = 0
Finally, solve this quadratic equation to find the value of 'a'.
The driver spends 4.7 times more time accelerating as decelerating. The average speed is half the speed limit during each interval. Therefore he travels 4.7 times farther while accelerating. The accelerating llength travelled is
(4.7/5.7)*3.31 = ?