Math / Algebra

Question21 22 23 24 25 areg is in a car at the top of a roller-coaster ride. The distance, $d$ , of the car from the ground as the car descends is letermined by the equation $d=144-16 t^{2}$ , where $t$ is the number of seconds it takes the car to travel down to each oint on the ride. For which interval of time is Greg's car moving in the air? $0<t \leq 3$ $-3<t \leq 3$ $t>3$ $0<t<3$ hark this and return Save and Exit Next Submit

Studdy Solution

STEP 1

1. The equation $d = 144 - 16t^2$ represents the distance from the ground, where $d \geq 0$ while the car is in the air.
2. The car starts at the top of the ride, and $t$ represents time in seconds.
3. We need to find the interval of time $t$ for which the car is above the ground, i.e., $d > 0$ .

STEP 2

1. Set the distance equation $d = 144 - 16t^2$ greater than zero to find when the car is in the air.
2. Solve the inequality for $t$ .
3. Determine the correct interval from the given options.

STEP 3

Set the equation for distance $d$ greater than zero to find when the car is in the air:
$144 - 16t^2 > 0$

STEP 4

Solve the inequality $144 - 16t^2 > 0$ . First, isolate $t^2$ by adding $16t^2$ to both sides:
$144 > 16t^2$

STEP 5

Divide both sides by 16 to solve for $t^2$ :
$\frac{144}{16} > t^2$ $9 > t^2$

STEP 6

Take the square root of both sides, remembering to consider both the positive and negative roots:
$\sqrt{9} > |t|$ $3 > |t|$
This implies:
$-3 < t < 3$

STEP 7

Determine the correct interval from the given options. Since $-3 < t < 3$ and $t$ must be positive as time cannot be negative, we refine the interval to:
$0 < t < 3$
The correct interval of time for which Greg's car is moving in the air is:
$\boxed{0 < t < 3}$

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Studdy Solution

STEP 1

1. The equation $d = 144 - 16t^2$ represents the distance from the ground, where $d \geq 0$ while the car is in the air.
2. The car starts at the top of the ride, and $t$ represents time in seconds.
3. We need to find the interval of time $t$ for which the car is above the ground, i.e., $d > 0$ .

STEP 2

1. Set the distance equation $d = 144 - 16t^2$ greater than zero to find when the car is in the air.
2. Solve the inequality for $t$ .
3. Determine the correct interval from the given options.

STEP 3

Set the equation for distance $d$ greater than zero to find when the car is in the air:
$144 - 16t^2 > 0$

STEP 4

Solve the inequality $144 - 16t^2 > 0$ . First, isolate $t^2$ by adding $16t^2$ to both sides:
$144 > 16t^2$

STEP 5

Divide both sides by 16 to solve for $t^2$ :
$\frac{144}{16} > t^2$ $9 > t^2$

STEP 6

Take the square root of both sides, remembering to consider both the positive and negative roots:
$\sqrt{9} > |t|$ $3 > |t|$
This implies:
$-3 < t < 3$

STEP 7

Determine the correct interval from the given options. Since $-3 < t < 3$ and $t$ must be positive as time cannot be negative, we refine the interval to:
$0 < t < 3$
The correct interval of time for which Greg's car is moving in the air is:
$\boxed{0 < t < 3}$

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Studdy Solution

STEP 1

STEP 2

1. Set the distance equation d=144−16t2 d = 144 - 16t^2 d=144−16t2 greater than zero to find when the car is in the air.2. Solve the inequality for t t t.3. Determine the correct interval from the given options.

STEP 3

Set the equation for distance d d d greater than zero to find when the car is in the air:144−16t2>0 144 - 16t^2 > 0 144−16t2>0

STEP 4

Solve the inequality 144−16t2>0 144 - 16t^2 > 0 144−16t2>0. First, isolate t2 t^2 t2 by adding 16t2 16t^2 16t2 to both sides:144>16t2 144 > 16t^2 144>16t2

STEP 5

Divide both sides by 16 to solve for t2 t^2 t2:14416>t2 \frac{144}{16} > t^2 16144​>t2 9>t2 9 > t^2 9>t2

STEP 6

Take the square root of both sides, remembering to consider both the positive and negative roots:9>∣t∣ \sqrt{9} > |t| 9​>∣t∣ 3>∣t∣ 3 > |t| 3>∣t∣This implies:−3<t<3 -3 < t < 3 −3<t<3

STEP 7

Determine the correct interval from the given options. Since −3<t<3 -3 < t < 3 −3<t<3 and t t t must be positive as time cannot be negative, we refine the interval to:0<t<3 0 < t < 3 0<t<3 The correct interval of time for which Greg's car is moving in the air is:0<t<3 \boxed{0 < t < 3} 0<t<3​

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1. Set the distance equation $d = 144 - 16t^2$ greater than zero to find when the car is in the air.
2. Solve the inequality for $t$ .
3. Determine the correct interval from the given options.

Set the equation for distance $d$ greater than zero to find when the car is in the air:
$144 - 16t^2 > 0$

Solve the inequality $144 - 16t^2 > 0$ . First, isolate $t^2$ by adding $16t^2$ to both sides:
$144 > 16t^2$

Divide both sides by 16 to solve for $t^2$ :
$\frac{144}{16} > t^2$ $9 > t^2$

Take the square root of both sides, remembering to consider both the positive and negative roots:
$\sqrt{9} > |t|$ $3 > |t|$
This implies:
$-3 < t < 3$

Determine the correct interval from the given options. Since $-3 < t < 3$ and $t$ must be positive as time cannot be negative, we refine the interval to:
$0 < t < 3$
The correct interval of time for which Greg's car is moving in the air is:
$\boxed{0 < t < 3}$