A person starting from rest acquires a speed of 20m/s in 10s . Then tr

1.	A person starting from rest acquires a speed of 20m/s in 10s . Then travel uniformly for 5s. The total distance travelled after 15 s isNB- IF YOU DON'T KNOW ANSWER( please don't give answer) (if you will give wrong answer I will report you) don't copy paste from google because I have already searched and didn't got answer.
Answer» Solution: Given, PERSON started from rest Acquires a speed of 20 m/s in 10s After travelled uniformly for 5s To FIND, TOTAL distance traveled after 15 s ________________________________ Firstly FINDING acceleration a = (v - u)/t Where a : acceleration v : FINAL velocity u : initial velocity t : time taken Here , final velocity (u) = 20m/s & initial velocity (u) = 0 m/s because the person starts from rest → a = 20 - 0/10 → a = 20/10 → acceleration = 2 m/s² Now , finding distance travelled in 15 s by using 1st equation of motion. S = ut + 1/2 at² Where s : distance u : initial velocity t : time a : acceleration → S = 0(15) + 1/2 (2)(15)² → S = 0 + 225 → S = 225 m Hence , distance travelled in 15s = 225 m

A person starting from rest acquires a speed of 20m/s in 10s . Then travel uniformly for 5s. The total distance travelled after 15 s isNB- IF YOU DON'T KNOW ANSWER( please don't give answer) (if you will give wrong answer I will report you) don't copy paste from google because I have already searched and didn't got answer.

Answer»

Solution:

Given,

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Firstly FINDING acceleration

a = (v - u)/t

Where

Here , final velocity (u) = 20m/s & initial velocity (u) = 0 m/s because the person starts from rest

→ a = 20 - 0/10

→ a = 20/10

→ acceleration = 2 m/s²

Now , finding distance travelled in 15 s by using 1st equation of motion.

S = ut + 1/2 at²

Where

→ S = 0(15) + 1/2 (2)(15)²

→ S = 0 + 225

→ S = 225 m

Hence , distance travelled in 15s = 225 m