a body moving with an inertia velocity of 30m/s accelerate uniformly at a rate of 10m/s until it attends a velocity of 50m/s.what is the distance covered during this period?
it takes (50-30)/10 = 2 seconds to reach the final velocity.
s = 30t + 1/2 t^2
Now just plug in t=2.
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A body moving with an intertia velocity of 30/s accelerate uniformly at a rete of 10m/s until it attend a velocity of 50m/s . What is the distance covered during the period?
Very good
What is heat period
It takes 50-30÷10=2seconds to reach final velocity
s=ut+1/2at^2
Then u will input the values
s=30t+1/2t^2
Input 2 for t and solve
To find the distance covered during this period, we need to use the equation:
distance = (initial velocity * time) + (1/2 * acceleration * time^2)
Given:
Initial velocity (u) = 30 m/s
Acceleration (a) = 10 m/s^2
Final velocity (v) = 50 m/s
First, we need to find the time taken to reach the final velocity. We can use the formula:
(v - u) = a * t
Rearranging the equation, we get:
t = (v - u) / a
Substituting the given values:
t = (50 - 30) / 10
t = 20 / 10
t = 2 seconds
Now, we can substitute the values of initial velocity, acceleration, and time into the distance equation:
distance = (30 * 2) + (1/2 * 10 * 2^2)
distance = (60) + (1/2 * 10 * 4)
distance = 60 + (1/2 * 40)
distance = 60 + 20
distance = 80 meters
Therefore, the distance covered during this period is 80 meters.