Question

35. A particle moves 200 cm in the first 2s and 220
cm in the next 4s with uniform deceleration.
The velocity of the particle at the end of 7s is
a) 12 cms-1
b) 11 cms-1
c) 10cms-1
d)5 cms-1​

Answer 1

Answer:

d is the answer

please mark it as a brainliest

please

as 1+1=2

Answer 2

Answer:

Correct answer -- (C) 10 cm/sec

Explanation -

Follow this method

Let 'u' be the initial velocity and 'a' the acceleration.

So we have the distance formula

s = ut + 1/2 at^2

In first 2 seconds,

s = 200

Put the values in the formula.

200 = u x 2 + 1/2 x a x (2)^2

200 = 2u + 1/2 x 4 x a

200 = 2u + 2a

100 = u + a -----(I)

In next 4 seconds

Let the distance traveled be 'y'

Time = 2+4 = 6 sec

So according to the formula ,

y = u x 6 + 1/2a x (6)^2

y = 6u + 1/2 x 36 x a

y = 6u + 18a ------(ii)

Now we know that 220 cm was traveled in between 2 sec - 6 sec

y - 200 = 220

y = 420

We know y = 6u + 18a (from [ii])

So, 6u + 18a = 420

u + 3a = 70 ------(iii)

Equating (I) and (iii)

-2a = 30

a = -15 cm/s^2

u = 100 - (-15)

u = 100 + 15 = 115 cm/sec

Now we know v = u + at

We have to find the "v" after 7th second

So v = 115 + (-15) x 7

v = 115 - 105

v = 10 cm/sec