Question

a partcle starts its motion from origin with velocity 4 m/s in positive x direction.lts acceleration is related with position as a=(2x+2). find magnitude of velocity of particle at x=4.

Answer 1

initial speed of particle , u = 4m/s

acceleration is varies as a = (2x + 2)

we know, a = v dv/dx, use it above .

so, v dv/dx = 2x + 2

or, $\int\limits^v_4{v}\,dv=\int\limits^4_0{(2x+2)}\,dx$

or, $\left[\frac{v^2}{2}\right]^v_4=\left[x^2+2x\right]^4_0$

or, (v² - 4²)/2 = [4² + 2 × 4 - 0]

or, (v² - 16) = 2 (16 + 8) = 48

or, v² = 48 + 16 = 64

or, v = 8 m/s

hence , magnitude of velocity of at x =4 is 8m/s

Answer 2

Answer:

Hey mate here is ur answer to the question... Hope it helps

Explanation:

initial speed of particle , u = 4m/s

acceleration is varies as a = (2x + 2)

we know, a = v dv/dx, use it above .

so, v dv/dx = 2x + 2

or, \int\limits^v_4{v}\,dv=\int\limits^4_0{(2x+2)}\,dx

4

∫

v

vdv=

0

∫

4

(2x+2)dx

or, \left[\frac{v^2}{2}\right]^v_4=\left[x^2+2x\right]^4_0[

2

v

2

]

4

v

=[x

2

+2x]

0

4

or, (v² - 4²)/2 = [4² + 2 × 4 - 0]

or, (v² - 16) = 2 (16 + 8) = 48

or, v² = 48 + 16 = 64

or, v = 8 m/s

hence , magnitude of velocity of at x =4 is 8m/s