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.

Give them answer

How pls solve

No

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.

A body moving with anlnitial velocity of 30ms-1 accelerates uniformly at a rate of 10ms-1 until it atlains a velocity of 50m-1. What is the distance covered during this period ?V2= U2+2as makes the subject