# Math Help - Antiderivatives

1. ## Antiderivatives

Near the earth's surface every particle experiences a constant acceleration of 32 ft/sec^2 directed toward the center of the earth; if "up" is considered the positive direction, then this is an acceleration of -32ft/sec^2. Consider a rock thrown straight up in the air from the ground. Let's denote time t=0 as the time when the rock reaches 12 feet up in the air at (upward) velocity of 4 feet per second. Fill in the following with numbers that differ from the correct value by no more than .1 units.

The rock left the ground (was thrown up in the air) at time t= ? seconds, with initial velocity of ? feet per second. It will fall back to the ground at time t= ? second after reaching the maximum height of ? feet at time t= ? seconds.

2. Just in case a picture helps to derive the motion equations you need...

Start by defining v as the rate of change of distance s with respect to time t, and a as the rate of change of v...

... where straight continuous lines differentiate downwards (integrate up) with respect to t.

Integrating a with respect to t, with u the traditional constant of integration, interpreted as intitial velocity...

Then integrate again, but this time we can do without the constant, which we are choosing as zero (see why)...

Either start subbing in the given values (v = 4, and a = -32) and see where it takes you, or use the middle row to get u as the expression v - at, and sub that expression into the top row and then plug in the values. You should be able to solve for t, being the first question.

Hope that helps.

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3. Originally Posted by Velvet Love
Near the earth's surface every particle experiences a constant acceleration of 32 ft/sec^2 directed toward the center of the earth; if "up" is considered the positive direction, then this is an acceleration of -32ft/sec^2. Consider a rock thrown straight up in the air from the ground. Let's denote time t=0 as the time when the rock reaches 12 feet up in the air at (upward) velocity of 4 feet per second. Fill in the following with numbers that differ from the correct value by no more than .1 units.

The rock left the ground (was thrown up in the air) at time t= ? seconds, with initial velocity of ? feet per second. It will fall back to the ground at time t= ? second after reaching the maximum height of ? feet at time t= ? seconds.
So dv/dt= -32 which means $dx/dt= v(t)= -32t+ v_0$ where $v_0$ is the speed at t= 0. Integrating again, $x(t)= -16t^2+ v_0t+ x_0$ where $x_0$ is the height at t= 0. You are told that, at t= 0, "the rock reaches 12 feet up in the air (so $x_0= 12$) at (upward) velocity of 4 feet per second (so $v_0= 4$).
When the rock "left the ground" or is "back to the ground", x= 0. Setting x(t)= 0 gives a quadratic equation for t which will have two solutions. The smaller t will be when the rock left the ground and the larger t will be when it hits the ground again. It's intial velocity will be v(t) where t is that smaller value. You can find the maximum height either by setting v(t)= 0 and then determining the height at that t or by completing the square in the formula for x(t).