a car starts from rest and is accelerated uniformly and the rest of 2m/s2 for 6seconds it then maintaining a constant speed for half a minutes.The brake are then applied and then vehicles unifomly restarted to rest in 5seconds.find maxmum speed in km/h and the total distance covered in metres
B. Area of triangle 1(d1) = 1/2 × 6 × 12 = 200
Area of rectangle ( d2) = 30 × 12 = 360
Area of triangle 2 (d3) = 1/2 × 5 × 12 = 30
Total distance (d) = (d1 + d2 + d3)m
= (200 + 360 + 30)m
= 426m
A.Maximum speed (V) = a × t
= 2m/s^2 × 6s
= 12m/s
= 0.012km/s × 3600s/h
V = 43.2km/h
a. V = a*t = 2 * 6 = 12 m/s.
V = 0.012km/s * 3600s/h = 43.2 km/h.
b. d1 = 0.5a*t^2 = 0.5*2*6^2 = 36 m.
d2 = V*t = 12 * 30 = 360 m.
V = Vo + a*t = 0.
12 + a*6 = 0, a = -2 m/s^2
d3 = Vo*t + 0.5a*t^2.
d3 = 12*6 - 1*5^2 = 47 m.
d = d1+d2+d3 = Total distance.
Correction: After the brakes were applied, V = Vo + a*t = 0.
12 + a*5 = 0, a = -2.4 m/s^2.
d3 = Vo*t + 0.5a*t^2.
d3 = 12*5 - 1.2*5^2 = 30 m.
d = d1+d2+d3 = 36 + 360 + 30 = 426 m. = Total distance covered.
To find the maximum speed in km/h and the total distance covered, we need to break down the various stages of the car's motion and calculate the results for each stage separately.
1. Acceleration phase:
The car starts from rest and accelerates uniformly at 2 m/s^2 for 6 seconds.
- Firstly, we need to find the final velocity (V) after this acceleration phase. We can use the formula:
V = u + at
Where:
- u is the initial velocity (0 m/s for a car at rest)
- a is the acceleration (2 m/s^2)
- t is the time (6 seconds)
Plugging in these values:
V = 0 + (2 * 6)
V = 12 m/s
- Now, let's convert the final velocity from m/s to km/h:
To convert m/s to km/h, we need to multiply by 3.6 (since 1 km = 1000 m and 1 hour = 3600 seconds).
Therefore, the final velocity in km/h is:
V_km/h = V * 3.6
V_km/h = 12 * 3.6
V_km/h = 43.2 km/h
So, the maximum speed reached by the car is 43.2 km/h during the acceleration phase.
2. Constant speed phase:
The car maintains a constant speed for half a minute, which is 30 seconds.
- During this phase, since the speed is constant, there is no acceleration. Thus, the velocity remains unchanged.
3. Deceleration phase:
The brakes are then applied, and the vehicle uniformly restarts to rest in 5 seconds.
- Similar to the acceleration phase, we can use the formula:
V = u + at
Where:
- u is the initial velocity (the constant speed reached in the previous phase)
- a is the acceleration (negative since it's deceleration)
- t is the time (5 seconds)
Since the car is slowing down and coming to rest, the final velocity (V) is 0 m/s. Plugging in these values:
0 = V + (-a * t)
0 = V - (a * 5)
Rearranging the equation to solve for V:
V = a * 5
Since the initial velocity (u) is the constant speed reached during the constant speed phase, let's call it V_constant.
Therefore:
V = V_constant - (a * 5)
- Now, let's substitute the values:
V = 43.2 - (a * 5)
To solve for a, we need one more piece of information. We can use the fact that the time it takes to decelerate is equal to the sum of the acceleration and deceleration times, i.e., 6 + 5 seconds.
Therefore, plugging in the values:
11 = 43.2 - (a * 5)
Rearranging the equation to solve for a:
a = (43.2 - 11) / 5
a = 6.24 m/s^2
Now, let's use this value of a to find the initial velocity (V_constant):
V = V_constant - (6.24 * 5)
0 = V_constant - 31.2
V_constant = 31.2 m/s
- Again, let's convert the final velocity from m/s to km/h:
V_constant_km/h = V_constant * 3.6
V_constant_km/h = 31.2 * 3.6
V_constant_km/h = 112.32 km/h
So, the maximum speed reached during the deceleration phase is 112.32 km/h.
To find the total distance covered, we need to calculate the distances covered in each stage and sum them up.
1. Distance during the acceleration phase:
To find the distance covered during uniformly accelerated motion, we can use the formula:
S = ut + (1/2)at^2
Where:
- u is the initial velocity (0 m/s)
- a is the acceleration (2 m/s^2)
- t is the time (6 seconds)
Plugging in these values:
S = 0 + (1/2) * 2 * (6^2)
S = 0 + 6 * 12
S = 72 meters
2. Distance during the constant speed phase:
Since the speed is constant, the distance covered can be found using the formula:
S = V * t
Where:
- V is the constant velocity (12 m/s, since it was the final velocity after the acceleration phase)
- t is the time (30 seconds)
Plugging in these values:
S = 12 * 30
S = 360 meters
3. Distance during the deceleration phase:
The distance covered during uniformly decelerated motion can be found using the same formula as in the acceleration phase:
S = ut + (1/2)at^2
But this time, since the final velocity is 0 m/s, the formula simplifies to:
S = (1/2)at^2
Where:
- a is the deceleration (6.24 m/s^2)
- t is the time (5 seconds)
Plugging in these values:
S = (1/2) * 6.24 * (5^2)
S = 15.6 * 25
S = 390 meters
Now, let's sum up the distances covered in each phase:
Total distance = Distance during acceleration phase + Distance during constant speed phase + Distance during deceleration phase
= 72 meters + 360 meters + 390 meters
= 822 meters
Therefore, the car covers a total distance of 822 meters during its motion.