**Q.4.21 A particle starts from the origin at t = 0 s with a velocity of 10.0 ? m/s and moves in the x-y plane with a constant acceleration of (8.0 ? + 2.0 ?) m s^-2.

(a) At what time is the x- coordinates of the particle 16 m? What is the y-coordinate of the particle at that time?

(b)** What is the speed of the particle at the time?

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Vishal Baghel | Contributor-Level 10

5 months ago

Ans.4.21:

(a) Velocity , $\overset{?}{v}$ = 10.0 ? m/s

Acceleration, $\overset{?}{a}$ = (8.0 ? + 2.0 ?) m s-2

We know $\overset{?}{a}$ = $\frac{d \overset{?}{v}}{d t}$ = 8.0 ? + 2.0 ?

$\overset{?}{d v}$ = (8.0 ? + 2.0 ?)dt

Integrating both sides we get $\overset{?}{v}$ (t) = 8.0t ? + 2.0t ? + $\overset{?}{u}$ ,

Where, $\overset{?}{u} =$ velocity vector of the particle at t =0

$\overset{?}{v} =$ velocity vector of the particle at time t

But $\overset{?}{v}$ = $\frac{\underset{d r}{?}}{d t}$

$\overset{?}{d r}$ = $\overset{?}{v}$ dt

= (8.0t ? + 2.0t ? + $\overset{?}{u}$ )dt

Integrating both sides with the condition at t = 0, r =0 and at t =t, r = r

$\overset{?}{r} =$ $\overset{?}{u}$ t + ½ $*$ 8.0 t2 ? + ½ $*$ 2.0 $*$ t2 ? = $\overset{?}{u}$ t + 4.0 t2 ? + t2 ?

Substituting the value of $\overset{?}{u}$ , we get

$\overset{?}{r} =$ ( 10.0 ?)t + 4.0 t2 ? + t2 ? . This equation can be expressed as

x ? + y ? = 4.0 t2 ? + ( 10.0t + t2) ?

Since the motion of the particle is confined to the x-y plane, on equating the coefficients of ? and ?, we get

x = 4.0 t2 and y = 10.0t + t2

t = $\sqrt{x / 4}$

(a) When x = 16m, t = 2 s, y = 24m

(b) Velocity of the particle

$\overset{?}{v}$ (t) = 8.0t ? + 2.0t ? + $\overset{?}{u}$

At t = 2 s,

$\overset{?}{v}$ (t) = 8.0 $*$ 2 ? + 2.0 $* 2$ ? + 10 $*$ ? = 16 ? + 14 ?

The magnitude of $\overset{?}{v}$ (t) is given by

$|\overset{?}{v}|$ = ( 162 + 142)1/2 = 21.26 m/s

<table border="1"><tbody><tr><td>Ans.4.21:(a) Velocity , <math><mover accent="true"><mrow><mrow><mi>v</mi></mrow></mrow><mo>?</mo></mover></math> = 10.0 ? m/sAcceleration, <math><mover accent="true"><mrow><mrow><mi>a</mi></mrow></mrow><mo>?</mo></mover></math> = (8.0 ? + 2.0 ?) m s-2We know <math><mover accent="true"><mrow><mrow><mi>a</mi></mrow></mrow><mo>?</mo></mover></math> = <math><mfrac><mrow><mrow><mi>d</mi><mover accent="true"><mrow><mrow><mi>v</mi></mrow></mrow><mo>?</mo></mover></mrow></mrow><mrow><mrow><mi>d</mi><mi>t</mi></mrow></mrow></mfrac></math> = 8.0 ? + 2.0 ?<math><mover accent="true"><mrow><mrow><mi>d</mi><mi>v</mi></mrow></mrow><mo>?</mo></mover></math> = (8.0 ? + 2.0 ?)dtIntegrating both sides we get <math><mover accent="true"><mrow><mrow><mi>v</mi></mrow></mrow><mo>?</mo></mover></math> (t) = 8.0t ? + 2.0t ? + <math><mover accent="true"><mrow><mrow><mi>u</mi></mrow></mrow><mo>?</mo></mover></math> ,Where, <math><mover accent="true"><mrow><mrow><mi>u</mi></mrow></mrow><mo>?</mo></mover><mo>=</mo><mi></mi></math> velocity vector of the particle at t =0<math><mover accent="true"><mrow><mrow><mi>v</mi></mrow></mrow><mo>?</mo></mover><mo>=</mo><mi></mi></math> velocity vector of the particle at time tBut <math><mover accent="true"><mrow><mrow><mi>v</mi></mrow></mrow><mo>?</mo></mover></math> = <math><mfrac><mrow><mrow><mrow><munder accentunder="false"><mo>?</mo><mrow><mrow><mi>d</mi><mi>r</mi></mrow></mrow></munder></mrow></mrow></mrow><mrow><mrow><mi>d</mi><mi>t</mi></mrow></mrow></mfrac><mi></mi></math><math><mover accent="true"><mrow><mrow><mi>d</mi><mi>r</mi></mrow></mrow><mo>?</mo></mover></math> = <math><mover accent="true"><mrow><mrow><mi>v</mi></mrow></mrow><mo>?</mo></mover></math> dt= (8.0t ? + 2.0t ? + <math><mover accent="true"><mrow><mrow><mi>u</mi></mrow></mrow><mo>?</mo></mover></math> )dtIntegrating both sides with the condition at t = 0, r =0 and at t =t, r = r<math><mover accent="true"><mrow><mrow><mi>r</mi></mrow></mrow><mo>?</mo></mover><mo>=</mo></math> <math><mover accent="true"><mrow><mrow><mi></mi><mi>u</mi></mrow></mrow><mo>?</mo></mover></math> t + ½ <math><mo>×</mo></math> 8.0 t2 ? + ½ <math><mo>×</mo></math> 2.0 <math><mo>×</mo></math> t2 ? = <math><mover accent="true"><mrow><mrow><mi></mi><mi>u</mi></mrow></mrow><mo>?</mo></mover></math> t + 4.0 t2 ? + t2 ?Substituting the value of <math><mover accent="true"><mrow><mrow><mi>u</mi></mrow></mrow><mo>?</mo></mover></math> , we get<math><mover accent="true"><mrow><mrow><mi>r</mi></mrow></mrow><mo>?</mo></mover><mo>=</mo></math> ( 10.0 ?)t + 4.0 t2 ? + t2 ? . This equation can be expressed asx ? + y ? = 4.0 t2 ? + ( 10.0t + t2) ?Since the motion of the particle is confined to the x-y plane, on equating the coefficients of ? and ?, we getx = 4.0 t2 and y = 10.0t + t2t = <math><msqrt><mrow><mi>x</mi><mo>/</mo><mn>4</mn></mrow></msqrt></math>(a) When x = 16m, t = 2 s, y = 24m(b) Velocity of the particle<math><mover accent="true"><mrow><mrow><mi>v</mi></mrow></mrow><mo>?</mo></mover></math> (t) = 8.0t ? + 2.0t ? + <math><mover accent="true"><mrow><mrow><mi>u</mi></mrow></mrow><mo>?</mo></mover></math>At t = 2 s,<math><mover accent="true"><mrow><mrow><mi>v</mi></mrow></mrow><mo>?</mo></mover></math> (t) = 8.0 <math><mo>×</mo><mi></mi></math> 2 ? + 2.0 <math><mo>×</mo><mn>2</mn></math> ? + 10 <math><mo>×</mo><mi></mi></math> ? = 16 ? + 14 ?The magnitude of <math><mover accent="true"><mrow><mrow><mi>v</mi></mrow></mrow><mo>?</mo></mover></math> (t) is given by<math><mfenced open="|" close="|" separators="|"><mrow><mrow><mi mathvariant="normal"></mi><mover accent="true"><mrow><mrow><mi>v</mi></mrow></mrow><mo>?</mo></mover></mrow></mrow></mfenced></math> = ( 162 + 142)1/2 = 21.26 m/s</td></tr></tbody></table>

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