5.11 A truck starts from rest and accelerates uniformly at 2.0 m s-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11s? (Neglect air resistance.)

The acceleration of the truck, a = 2 m/s², t = 10 s, Initial velocity, u = 0, from the equation v = u + at, we get v = 2 $\times 10$ = 20 m/s

At time t = 11 s, the horizontal component of the velocity Vx remains unchanged due to no air resistance. Hence Vx = 20 m/s

The vertical component of the velocity Vy is expressed as

Vy = u + $a_y$ t

Here t =11-10 = 1 s,  $a_y=a=g=$ 10 m/s², u = 0

Vy = 10 m/s

The resultant velocity V is given by V = ( $V_x^2$ + $V_y^2$ )1/2 = 22.36 m/s

$\tan ??$ =Vy/Vx=10/20,  $?=$ 26.57 $°$ w.r.t. horizontal

 

<p>The acceleration of the truck, a = 2 m/s², t = 10 s, Initial velocity, u = 0, from the equation v = u + at, we get v = 2 <span title="Click to copy mathml"><math><mo>×</mo><mn>10</mn></math></span>&nbsp;= 20 m/s</p><p>At time t = 11 s, the horizontal component of the velocity Vx remains unchanged due to no air resistance. Hence Vx = 20 m/s</p><p>The vertical component of the velocity Vy is expressed as</p><p>Vy = u +&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mi>y</mi><mi></mi></mrow></mrow></msub></math></span>&nbsp;t</p><p>Here t =11-10 = 1 s, &nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mi>y</mi><mi></mi></mrow></mrow></msub><mo>=</mo><mi>a</mi><mo>=</mo><mi>g</mi><mo>=</mo></math></span> 10 m/s², u = 0</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1728295732phpIvjcD7_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1728295732phpIvjcD7_480x360.jpeg"6 alt="" width="327" height="230"></picture></div></div><p>Vy = 10 m/s</p><p>The resultant velocity V is given by V = (&nbsp;<span title="Click to copy mathml"><math><msubsup><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mi>x</mi></mrow></mrow><mrow><mrow><mn>2</mn><mi></mi></mrow></mrow></msubsup></math></span>&nbsp;+&nbsp;<span title="Click to copy mathml"><math><msubsup><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mi>y</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msubsup></math></span>&nbsp;)1/2 = 22.36 m/s</p><p><span title="Click to copy mathml"><math><mrow><mrow><mi mathvariant="normal">tan</mi></mrow><mo>? </mo><mrow><mi>? </mi></mrow></mrow></math></span>&nbsp;=Vy/Vx=10/20, &nbsp;<span title="Click to copy mathml"><math><mi>? </mi><mo>=</mo></math></span>&nbsp;26.57&nbsp;<span title="Click to copy mathml"><math><mo>°</mo></math></span>&nbsp;w.r.t. horizontal</p><p>&nbsp;</p>

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