Question

a particle is moving is positive x-direction with initial velocity of 10m/s and uniform retardation such that it reaches the initial position after 10s. the distance traversed by the particle in 6s is​

Answer 1

Given :

1) initial Velocity= 10m/s

2) It moves with uniform retardation such that it reaches the initial position after 10 sec.i.e

S = 0

and t = 10 sec.

To find

The distance traveled by the particle in 6 sec.

$\purple{\boxed{\large{\bold{Formula's}}}}$

Kinematic equations for uniformly accelerated motion.

$\bf v=u+at$

$\bf s=ut+ \frac{1}{2} a{t}^{2}$

$\bf {v}^{2}={u}^{2}+2as$

and $\bf s_{nth} = u+ \frac{a}{2} (2n-1)$

Solution :

• Case 1)

We have

$\sf u=10m/s$

ΔS = 0

t = 10 sec

By equation of motion,

Δ $\sf{S=ut+\frac{1}{2}a{t}^{2}}$

Put the given values

Δ $\sf{0=10 \times 10+ \frac{1}{2} \times a \times10^{2}}$

$\sf{-100=\frac{1}{2} 100 a}$

$\sf{a=\frac{-100\times2}{100}}$

$\sf{a=-2ms^{-2}}$

• Case 2:

We have to find the distance traveled by the particle in 6 sec.

It is given that the particle reaches the Initial position after 10 sec. Thus

Particle come to rest after 5 seconds

Distance traveled in 5 sec

$\sf\:S_1=ut+\dfrac{1}{2}at^2$

$\sf\:S_1=10\times5+\dfrac{1}{2}\times-2\times25$

$\sf\:S_1=25m$

After 5s velocity becomes 0 and the particle starts to move in negative x-direction with the acceleration of 2.

Distance traveled in 6th second

$\sf\:S_2=0\times\:t+\dfrac{1}{2}\times2\times1^2$

$\sf\:S_2=1m$

Therefore,

Distance traveled in 6 seconds,

S  = Distance in 5 seconds + Distance traveled in 6th second

$\sf\:S=S_1+S_2$

$\sf\:S=25+1$

$\sf\:S=26m$

Hence, total distance travelled = 25+1= 26m

Answer 2

$\mathtt{\huge{\underline{\red{Question\:?}}}}$

✴ A particle is moving is positive x-direction with initial velocity of 10m/s and uniform retardation such that it reaches the initial position after 10s. The distance travelled by the particle in 6s is ?

$\mathtt{\huge{\underline{\green{Answer:-}}}}$

✏ The distance covered by the particle in time period of 6 sec is 26m.

$\mathtt{\huge{\underline{\purple{Solution:-}}}}$

Given :-

• The Initial Velocity (u) = 10m/s

• During the uniform retardation , the initial position in time period (t) of 10 sec & distance(s) is 0.

To find :-

• The distance covered by the particle in time period of 6 sec.

Solution :-

According to the question,

• The Initial Velocity (u) = 10m/s

• During the uniform retardation , the initial position in time period (t) of 10 sec & distance(s) is 0.

Using, S = ut + 1/2at²

S = ut + 1/2at²

☛ 0 = 10×10 + 1/2 × a × 10²

☛0 = 100 + 1/2×100×a

☛ 0 = 100 + 50a

☛ -100 = 50a

☛ a = -100/50

☛ a = -2m/s²

Now, the distance covered by the particle.

S1 = ut +1/2at²

☛ S1 = 10 × 5 + 1/2 × -2 × 25

☛ S1 = 50 + 1/2 ×-50

☛ S1 = 50 -25

☛ S1 = 25 m

We have to find the distance in 6th second.

S2 = ut +1/2at²

☛ S2 = 0×t + 1/2×2×1

☛ S2 = 0 + 1×1

☛ S2 = 1

✴ The distance covered by the particle in time period of 6 sec.

S total = S1 + S2

☛ S total = 25 +1

☛S total = 26m

The total distance covered by the particle is = 26m.

______________________________________