1. If an object is thrown straight up from the ground, at what velocity does it
ID: 2851453 • Letter: 1
Question
1. If an object is thrown straight up from the ground, at what velocity does it need to be thrown in order to reach a height of 50 ft?
2. I am playing mini golf, and I hit my ball at a speed of 3 ft/s. It rolls through the course with acceleration a(t) = 12t2 + 24t + 36 ft/s2. How far away is the ball when it stops accelerating?
3. What is the area bound between the curve y = x3 + 3x + 3 and the x-axis from x = 0 to x = 2?
4. What is the area beneath the curve y = x2 6x 5 and above the x-axis?
5. What is the area beneath the curve y = e3x + 3 from x = 1 to x = 4?
Explanation / Answer
1. initial height = 0 ft
displacement = 50 ft
accel due to gravity = -32 ft/s^2
final velocity = 0. this is velocity in the y direction (vertical) at the max height.
v^2 = vo^2 + 2ax
0 = vo^2 + 2(-32)(50)
vo = sqrt(3200)
vo = 56.57 ft/s
2.a(t)=-12t^2+24t+36
when a(t)=0 then 0=-12t^2+24t+36
t^2-2t-3=0
t=3
distance =speed*time
=3*3=9
3. Take the integral from 0 to 2 of y=x^3+3x+3 with respect to x.
It would be (1/4)x^4+(3/2)x^2+3x, evaluated from 0 to 2:
(1/4)(2^4)+(3/2)(2^2)+3(2)-(1/4)(0^4)+...
4+6+6=16.
4 Take the integral from 0 to 5 of y=-x^2-6x-5 with respect to x.
It would be (1/3)x^3-(6/2)x^2-5x, evaluated from 0 to 5:
=(1/3)(5^3)-(3)(5^2)-5(2)
=5/3
5.
Take the integral from -1 to 4 of y=e^3x+3 with respect to x.
It would be e^(3x+3)3, evaluated from -1 to 4:
=e^(15)3-e^0
=e^(45)-1
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