Question

If a particle having an initial velocity of 20ms-1, moves with a uniform acceleration of 10ms-2, then the distance covered by it when it attains a final velocity of 80 ms-1 is​

Answer 1

Answer:

the answer is 300 it is the right answer this question in my material

Explanation:

please follow

Answer 2

Given:

Initial velocity = u = 20m/s

Acceleration = a = 10ms-²

Final velocity = v = 80m/s

To find:

Distance travelled = s

Solution:

According to the third law of motion, for uniform acceleration, the formula used for the following question is

${v}^{2} - {u}^{2} = 2as$

where

-"v" is equal to the final velocity

"u" is equal to the initial velocity

"a" is equal to acceleration and

" S" is equal to displacement or distance.

Now, putting the values mentioned in the above question into the given equation, we get,

${80}^{2} - {20}^{2} = 2 \times 10 \times s$

S = 300m

Hence, the distance travelled by the particle is 300m