How Does Calculus Help Solve Motion Problems in Physics?

[gigi9]

Someone please help me with these problems below...I'm so confused...
Plz show me clearly how to begin...too...I'm not good with mixing calc and physics... Thanks very very much.
1) With what velocity must an arrow be shot upward in order to fall back to its starting point 10 seconds later? How high will it rise?
2)A woman standing on a bridge whrows a stone straight up. Exactly 5 seconds later the stone passes the woman on the way down, and 1 second after that it hits the water below. Find the initial velocity of the stone and the height of the bridge above the water.
3)How much time does a train traveling 144km/h take to stop if it ahs a constant negative acceleration of 4 m/s^2? How far does the train travel in this time?

 

[chroot]

Are you familiar with these equations?

$$v(t) = v_0 + a t$$

$$x(t) = x_0 + v_0 t + \frac{1}{2} a t^2$$

- Warren

 

[M98Ranger]

Hi there,

I understand that you are struggling with these problems and I am happy to help you. Calculus and physics can be challenging, but with practice and a clear understanding of the concepts, you can solve these problems easily.

1) For the first problem, we need to use the equation of motion under gravity: s(t) = ut + 1/2gt^2, where s(t) is the displacement at time t, u is the initial velocity, and g is the acceleration due to gravity (which is approximately 9.8 m/s^2 on Earth).

Since we want the arrow to fall back to its starting point, the displacement at time t = 10 seconds should be zero. Therefore, we can set s(10) = 0 and solve for u.

0 = u(10) + 1/2(9.8)(10)^2
0 = 10u + 490
-490 = 10u
u = -49 m/s

This means that the arrow must be shot upward with a velocity of 49 m/s. To find the height it will rise, we can use the equation v^2 = u^2 + 2gs, where v is the final velocity (which is 0 at the top of the trajectory) and s is the maximum height reached.

0^2 = (49)^2 + 2(9.8)s
0 = 2401 + 19.6s
-2401 = 19.6s
s = -122.45 m

Therefore, the arrow will rise to a height of 122.45 meters before falling back to its starting point.

2) For this problem, we need to use the same equation of motion under gravity. However, we need to consider the stone's motion in two parts: the upward motion and the downward motion.

For the upward motion, we know that the stone's initial velocity is the same as its final velocity when it reaches its maximum height (v = 0). Therefore, we can use the equation v = u + gt to find the initial velocity.

0 = u + (-9.8)(5)
0 = u - 49
u = 49 m/s

This means that the woman threw the stone with an initial velocity of 49 m/s. To find the height of the bridge, we can use the equation s = ut + 1/2gt^