Answer to Question #334775 in Math for Agl

Question #334775

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.

1
Expert's answer
2022-05-01T17:17:14-0400

5.


x(t)=x0+v0t+at22x(t)=x_0+v_0t+\dfrac{at^2}{2}

v(t)=v0+atv(t)=v_0+at

a)


x(2)=5.25+3(2)+4.15(2)22=19.55(m)x(2)=5.25+3(2)+\dfrac{4.15(2)^2}{2}=19.55(m)

b)


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

t=24.15st=\dfrac{2}{4.15} s

x(24.15)=5.25+3(24.15)+4.15(24.15)22x(\dfrac{2}{4.15} )=5.25+3(\dfrac{2}{4.15} )+\dfrac{4.15(\dfrac{2}{4.15} )^2}{2}

=29.78754.15(m)7.178(m)=\dfrac{29.7875}{4.15} (m)\approx7.178(m)

6.

a.


hmax=h0+v022gh_{max}=h_0+\dfrac{v_0^2}{2g}

hmax=64m+(4.9m/s)22(9.8m/s2)=65.225mh_{max}=64m+\dfrac{(4.9m/s)^2}{2(9.8m/s^2)}=65.225m

b.


h(t)=h0+v0tgt22h(t)=h_0+v_0t-\dfrac{gt^2}{2}

64+4.9t9.8t22=064+4.9t-\dfrac{9.8t^2}{2}=0

t2t644.9=0,t0t^2-t-\dfrac{64}{4.9}=0, t\ge0

t=1+1+4(644.9)2213.57(s)t=\dfrac{1+\sqrt{1+4(\dfrac{64}{4.9})^2}}{2}\approx13.57(s)

c.

vground=2ghmax|v_{ground}|=\sqrt{2gh_{max}}

vground=2(9.8m/s2)(65.225m)35.755m/s|v_{ground}|=\sqrt{2(9.8m/s^2)(65.225m)}\approx35.755m/s


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