Question

A ball bearing is placed on an inclined plane and begins to roll. The angle of elevation of the plane is θ. The distance (in meters) the ball bearing rolls in t seconds is s(t) = 4.9(sin θ)t2. (a) Determine the speed of the ball bearing after t seconds.

Answer 1

Answer:

Therefore  the velocity of the ball after t seconds is $9.8 sin\theta t$ m.

Step-by-step explanation:

Velocity: The ratio of distance to time.

Given that,

The distance the ball  bearing rolls in t seconds is

$S(t)=4.9 sin\theta t^2$

The first order derivative of distance of an object is the velocity of that object.

and the second order derivative of distance is acceleration.

$S(t)=4.9 sin\theta t^2$

Differentiating with respect of t

$\frac{d}{dt}S(t)=\frac{d}{dt}(4.9 sin\theta t^2)$

$\Rightarrow S'(t)= 4.9 sin\theta\frac{d}{dt}( t^2)$

$\Rightarrow S'(t)= 4.9 sin\theta(2t)$

$\Rightarrow S'(t)= 9.8 sin\theta t$

Therefore  the velocity of the ball after t seconds is $9.8 sin\theta t$ m.        

Answer 2

Speed of the ball bearing after t seconds = $9.8tsin\alpha$

It is given that the angle of elevation of the plane is α.

The distance function is given in the question

$S(t) = 4.9 sin \alpha* t^{2}$

What is velocity?

It is the rate of change of displacement with respect to time.

Speed will be the derivative of distance function with respect to time

Speed of the ball after t seconds

$\frac{d}{dt} S(t) = \frac{d}{dt} 4.9sin \alpha* t^{2} = 2*4.9 t sin\alpha = 9.8t sin \alpha$ ,sinα is constant here.

Thus, speed of the ball after t seconds = $9.8tsin\alpha$

To get more about distance, velocity and acceleration refer to the link,

https://brainly.com/question/25749514