A car accelerates from rest at 1.0 m/s^2 for 20s along a straight roa, it then moves at a constant speed for half an hour, it then deccelerates uniformly to a stop in 30s. Find the total distance covered by the car.
first leg:
a = 1
v = t + c
when t = 0 , v = 0 , thus c = 0
v = t
s = (1/2)t^2
so after 20 seconds,
s = (1/2)(400) = 200 m
v = 20 m/s (or 72 km/h )
second leg: at constant speed, no acceleration
v = 20 m/s
s = rate x time = 20(1800) = 36000 m
last leg:
a = ?
v = at + 20
when t = 30, v = 0, so
0 = 30a + 20
a = -1/15
s = (-1/15)t^2 + 20t
when t = 30
s = (-1/15)(900) + 30(20) = 540 m
so the total distance = 200 + 36000 + 540 m
= 36740 m
or
36.74 km
answer is reasonable, most of the trip is taken up by the 1/2 hr at a constant speed of 72 km/h which would be 36 km
Good
Kiss your left hand, say a weekday, put this in 15 other comment sections and he/she/ they will confess their feelings for you tomorrow.
Kiss your left hand, say a weekday, put this in 15 other comment sections and he/she/ they will confess their feelings for you tomorrow.
Well, let's break it down. We have three separate periods of motion: acceleration, constant speed, and deceleration.
1. First, let's calculate the distance covered during the acceleration phase using the formula: distance = (initial velocity * time) + (0.5 * acceleration * time^2).
Assuming the car starts from rest, the initial velocity is 0 m/s, the acceleration is 1.0 m/s^2, and the time is 20s. So, the distance covered during acceleration is:
distance_ac = 0.5 * 1.0 * (20)^2 = 200m.
2. Next, let's calculate the distance covered during the constant speed phase. The car is traveling at a constant speed for half an hour, which is 0.5 * 60 * 60 = 1800s.
Since the speed is constant, the distance covered is simply speed * time. Let's say the constant speed is v m/s. So, the distance covered during constant speed is:
distance_const = v * 1800m.
3. Finally, let's calculate the distance covered during the deceleration phase. The car decelerates uniformly to a stop in 30s.
Since the car is stopping, the final velocity is 0 m/s. Let's say the initial velocity during deceleration is u m/s. Using the formula: distance = (initial velocity * time) + (0.5 * acceleration * time^2), we can solve for u:
0 = u * 30 - 0.5 * 1.0 * (30)^2,
0 = 30u - 450,
30u = 450,
u = 15.
So, the distance covered during deceleration is:
distance_dec = (15 * 30) - (0.5 * 1.0 * (30)^2) = 225m.
Now, to find the total distance covered by the car, we add up the distances covered during each phase:
total_distance = distance_ac + distance_const + distance_dec.
And there you have it! The total distance covered by the car. Go ahead and plug in the values for the variables to get the exact answer. Just don't forget to calculate the constant speed phase using the value of v. Happy calculating!
To find the total distance covered by the car, we need to calculate the distance covered during each phase of the motion and then add them together.
Phase 1: Acceleration
The car accelerates from rest at 1.0 m/s^2 for 20 seconds. To find the distance covered during this phase, we can use the equation:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Here, the initial velocity is 0 m/s (rest), the acceleration is 1.0 m/s^2, and the time is 20 seconds. Plugging these values into the equation:
distance = (0 * 20) + (0.5 * 1.0 * (20^2))
= 0 + (0.5 * 1.0 * 400)
= 0 + 200
= 200 meters
So, the distance covered during the acceleration phase is 200 meters.
Phase 2: Constant Speed
The car moves at a constant speed for half an hour, which is equal to 0.5 * 60 = 30 minutes or 30 * 60 = 1800 seconds. Since the speed is constant, the distance covered during this phase is given by:
distance = speed * time
To calculate the distance, we need to know the speed at which the car is moving. Please provide the speed of the car during this phase.
Phase 3: Deceleration
During this phase, the car decelerates uniformly to a stop in 30 seconds. The deceleration is equal to the negative of the acceleration, which is -1.0 m/s^2. Using the same formula as in the acceleration phase, we can calculate the distance covered during this phase.
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Here, the initial velocity is the constant speed attained in phase 2 (which we need to know) and the deceleration is -1.0 m/s^2. The time for this phase is 30 seconds. Plugging these values into the equation will yield the distance covered in this phase.
Once we have the values for the speed of the car during the constant speed phase and the initial velocity in the deceleration phase, we can calculate the total distance covered by adding the distances from each phase.
first leg:
a = 1
v = t + c
when t = 0 , v = 0 , thus c = 0
v = t
s = (1/2)t^2
so after 20 seconds,
s = (1/2)(400) = 200 m
v = 20 m/s (or 72 km/h )
second leg: at constant speed, no acceleration
v = 20 m/s
s = rate x time = 20(1800) = 36000 m
last leg:
a = ?
v = at + 20
when t = 30, v = 0, so
0 = 30a + 20
a = -1/15
s = (-1/15)t^2 + 20t
when t = 30
s = (-1/15)(900) + 30(20) = 540 m
so the total distance = 200 + 36000 + 540 m
= 36740 m
or
36.74 km
answer is reasonable, most of the trip is taken up by the 1/2 hr at a constant speed of 72 km/h which would be 36 km