A car start from rest and is accelerated uniform at the rate of 2m/s'² for 6 seconds. If then maintain a constant speed for half a minutes. The breaks are then applied and the vehicle uniformly retardetered in 5 second. The maximum speed reached in km/h and the total distant covered in metres
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if a = constant
then
v = Vi + a t
x =Xi + Vi t + (1/2) a t^2
To find the maximum speed reached in km/h, we first need to calculate the final velocity after accelerating uniformly for 6 seconds.
Using the formula for uniformly accelerated motion:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time taken
At the start, the car is at rest, so the initial velocity (u) is 0 m/s.
Given:
Acceleration (a) = 2 m/s²
Time (t) = 6 seconds
v = 0 + 2 * 6
v = 12 m/s
Next, to convert the final velocity from m/s to km/h, we multiply by 3.6 (since 1 km/h = 1000 m/3600 s):
v_km/h = v * 3.6
v_km/h = 12 * 3.6
v_km/h = 43.2 km/h
Therefore, the maximum speed reached by the vehicle is 43.2 km/h.
Now, let's calculate the total distance covered.
In the first phase, the car undergoes uniform acceleration. The formula to calculate the distance covered during this phase is:
s = ut + (1/2) * a * t²
where:
s = distance
u = initial velocity
t = time taken
a = acceleration
Given:
Acceleration (a) = 2 m/s²
Time (t) = 6 seconds
s = 0 * 6 + (1/2) * 2 * (6^2)
s = 0 + 0.5 * 2 * 36
s = 0 + 1 * 36
s = 36 meters
In the second phase, the car maintains a constant speed for 30 seconds. The distance covered during this phase is:
s = v * t
where:
s = distance
v = velocity (which is constant during this phase)
t = time taken
Given:
Velocity (v) = 12 m/s
Time (t) = 30 seconds
s = 12 * 30
s = 360 meters
In the third phase, the car undergoes uniformly retarded motion. The formula to calculate the distance covered during this phase is the same as before:
s = ut + (1/2) * a * t²
In this case, the acceleration is negative (since the car is decelerating or slowing down). The magnitude of the acceleration will be the same as before (2 m/s²) but negative (-2 m/s²). The time (t) is given as 5 seconds.
s = 12 * 5 + (1/2) * (-2) * (5^2)
s = 60 + (-0.5) * 2 * 25
s = 60 + (-0.5) * 50
s = 60 + (-25)
s = 35 meters
Finally, the total distance covered is the sum of the distances covered in each phase:
total distance = 36 + 360 + 35
total distance = 431 meters
Therefore, the total distance covered by the vehicle is 431 meters.