Question

position of a particle moving along x-axis is given by x=t⁴-3t³+4t²+1 . where x is in meters and t is in seconds —

1) find position of particle at t=2sec.

2) find the displacement of particle on the time interval from t=0. to t=2sec​

Answer 1

Question

The position of a particle moving along X - axis is defined as x(t) = t⁴ - 3t³ + 4t² + 1 with respect to time.

• Find the position of the particle at t = 2s

• Find the displacement of the particle between t = 0s to t = 2s

Solution

• When t = 2s,x would be 9 units

• Displacement is 8 m

Given

The position of the particle is defined as :

$\sf \: x(t) = {t}^{4} - 3 {t}^{3} + 4 {t}^{2} + 1$

$\rule{300}{2}$

When t = 2s,the particle would be at :

$\sf \: x_2 = {2}^{4} - 3 \times 2 {}^{3} + 4 \times {2}^{2} + 1 \\ \\ \longrightarrow \: \sf \: x_2 = 16 - 3(8) + 16 + 1 \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf \:x_2 = 9 \: m }}$

The particle is 9 units away from the origin

$\rule{300}{2}$

When t = 0s,

$\boxed{ \boxed{ \sf \: x_1 = 1 \: m}}$

$\rule{300}{2}$

Displacement of the particle would be :

$\large{\sf \: \Delta{x} = x_2 - x_1 }\\ \\ \huge{\longrightarrow \: \underline{ \boxed{ \sf \: \Delta{x} = 8 \: m}}}$

$\rule{300}{2}$

$\rule{300}{2}$

Answer 2

Answer:

1. 9 m

2. 8 m

Explanation:

Given :

Displacement of particle :

x = t⁴ - 3 t³ + 4 t² + 1

1 . Displacement of particle at t = 2 sec

x ( 2 ) = 2⁴ - 3. 2³ + 4. 2² + 1

x ( 2 ) = 16 - 24 + 16 + 1

x ( 2 ) = 33 - 24

x ( 2 ) = 9 m

Therefore , positive of particle at t = 2 sec is 9 m .

2 . We have positive of particle at t = 2 sec is 9 m

Now position of particle at t = 0 sec .

x ( 0 ) = 0⁴ - 3. 0³ + 4. t² + 1

x ( 0 ) = 1 m

Displacement of particle between t = o sec to t = 2 sec .

= > 9 m - 1 m

= > 8 m

Hence we get answer.