a=v/t
a=38/10
a=-3.8 m/s^2
You put the negative in front of 3.8 because it decelerated.
Hope it helps and is correct :)
Whenever an object is in projectile motion, that is, it has 2-dimensional motion in the x and y axis, the resultant force on the object is in the y-direction.
This is because once the object has been projected, or the ball has been kicked in this case, there is no longer a force being applied on it in the x-direction. The air resistance is also neglected so the ball's final velocity in the x-direction is equal to its initial velocity in the x-direction.
However, the force of gravity cannot be neglected and causes the ball to come downwards. Therefore, after the ball has been projected, the net force on the ball is downwards, due to gravity.
Answer:
x(t) = -8sin2t
Explanation:
See the attachment for solution
From my solving, we can deduce that w² = 4, and thus, w = 2
Therefore, the general solution is
x(t) = c1 cos2t + c2 sin2t
Considering the final variable, we can conclude that
x(0) = 0
x'(0) = -8 m/s
The final solution, thus
x(t) = -8sin2t
For the answer to the question above, first find out the gradient.
<span>m = rise/run </span>
<span>=(y2-y1)/(x2-x1) </span>
<span>the x's and y's are the points given: "After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 km/h" </span>
<span>(x1,y1) = (3,53) </span>
<span>(x2,y2) = (6,62) </span>
<span>sub values back into the equation </span>
<span>m = (62-53)/(6-3) </span>
<span>m = 9/3 </span>
<span>m = 3 </span>
<span>now we use a point-slope form to find the the standard form </span>
<span>y-y1 = m(x-x1) </span>
<span>where x1 and y1 are any set of point given </span>
<span>y-53 = 3(x-3) </span>
<span>y-53 = 3x - 9 </span>
<span>y = 3x - 9 + 53 </span>
<span>y = 3x + 44 </span>
<span>y is the velocity of the car, x is the time.
</span>I hope this helps.
