Answer to Question #334775 in Math for Agl

Question

5. A cart is at x = 5.25m and time t = 0. The cart accelerates at

4.15m/s^2. If The speed of the cart at t=0 is 3m/s, find the position

of the cart at t =2s and also determine where the cart is when it

reaches a speed of 5m/s. (3 points)

6. A stone is thrown vertically up with an initial velocity of 4.9m/s from

the top of the building that is 64m high. On its way down, it misses

the top of the building and goes straight to the ground. Find:

(9 points)

a. Its maximum height relative to the ground

b. Its time of flight or the total time it is in air, and

c. Its velocity just before it reaches the ground.

Answer 1

5.

$$x(t)=x_0+v_0t+\dfrac{at^2}{2}$$

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

a)

$$x(2)=5.25+3(2)+\dfrac{4.15(2)^2}{2}=19.55(m)$$

b)

$$v(t)=3+4.15t=5$$

$$t=\dfrac{2}{4.15} s$$

$$x(\dfrac{2}{4.15} )=5.25+3(\dfrac{2}{4.15} )+\dfrac{4.15(\dfrac{2}{4.15} )^2}{2}$$

$$=\dfrac{29.7875}{4.15} (m)\approx7.178(m)$$

6.

a.

$$h_{max}=h_0+\dfrac{v_0^2}{2g}$$

$$h_{max}=64m+\dfrac{(4.9m/s)^2}{2(9.8m/s^2)}=65.225m$$

b.

$$h(t)=h_0+v_0t-\dfrac{gt^2}{2}$$

$$64+4.9t-\dfrac{9.8t^2}{2}=0$$

$$t^2-t-\dfrac{64}{4.9}=0, t\ge0$$

$$t=\dfrac{1+\sqrt{1+4(\dfrac{64}{4.9})^2}}{2}\approx13.57(s)$$

c.

$$|v_{ground}|=\sqrt{2gh_{max}}$$

$$|v_{ground}|=\sqrt{2(9.8m/s^2)(65.225m)}\approx35.755m/s$$