Question

A body travels 200cm in first 2sec and 220cm in the next 4sec. What will be the velocity at the end of the 7th second from the start?

Answer 1

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

Answer 2

Question: A body travels $200cm$ in first $2sec$ and $220cm$ in the next $4sec$. What will be the velocity at the end of the $7^{th}$ second from the start?

Solution: Let $u$ be the initial and $a$ be the acceleration of the body. Then, we have, using the equation of motion,

$x=ut+\frac{1}{2} at^{2}$

For first 2 seconds: $200=2u+\frac{1}{2} a(2)^{2}$

Distance travelled in first 6 seconds $= 200+220=420cm$

Therefore, for first 6 seconds: $420=6u+\frac{1}{2} a(6)^{2}$

Solving these equations, we get

$u=115cm/s,\:a=-15cm/s^{2}$

Velocity at the end of 7 seconds:

$v=u+at=155+(-15)\times7=10cm/s.$