Question

a body travels a distance of 2m in 2 seconds and 2.2 m in next 4 seconds what will be the velocity of body at the end of end of 7 second from start. please ans in detail . . . . . .. i will mark as brainliest

Answer 1

Let the initial velocity be ‘u’ and the constant acceleration be ‘a’.

Distance covered in first 2 seconds =2m

=> ut +½at²=2m

=>2u+2a=2m (t=2seconds)

=>u+a=1———(I)

Distance covered in last 4 seconds=2.2m

Total distance in 6 seconds =2.2+2=4.2m

=>ut+½at²=4.2m

=>6u+18a=4.2m (t=6 seconds)

=>u+3a=0.7-(ii)(dividing both sides by 6)

Now subtracting equation (ii) form (I),we get—

-2a=0.3

=>a= -0.15m/s²

Putting this value in equation (I), we have--

u=1.15m/s

Now , velocity at the end of 7th second = u+at = 1.15 + (-0.15*7)=1.15–1.05m/s = 0.1m/s

Hence, velocity at the end of 7th second is 0.1m/s or 10cm/s.

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Answer 2

Given :

• a body travels a distance of 2m in 2 seconds and 2.2 m in next 4 seconds

To find :

• the velocity of body at the end of end of 7 second from start.

Solution :

s = ut + ½ at² .......(formula)

let u = initial velocity

for 1st 2 seconds

2 = u(2) + ½ a(2)²

1 = u + a ...........(1)

in the next u second 2.2 m

t = 4 + 2 = 6 sec

s = 2 + 2.2 = 4.2 m

4.2 = 6u + ½ (a)(6)²

4.2 = 6u + 18a

1.4 = 2u + 6a ............(2)

from (1) and (2)

u = 1.15 m/s and a = -0.15 m/s²