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If a firework has an initial velocity of 235m/s and is in the air for 12 seconds with an acceleration of -10 m/s2 before exploding, how far did it go?

Sagot :

Sarah06109

Answer:

[tex]\boxed {\boxed {\sf 2100 \ meters}}[/tex]

Explanation:

We are asked to find the distance a firework travels.

We are given the initial velocity, acceleration, and time, but we don't know the final velocity. Therefore, we will use the following kinematic equation.

[tex]d=v_it+\frac{1}{2} at^2[/tex]

The initial velocity is 235 meters per second. The firework travels for 12 seconds. It has an acceleration of -10 meters per second squared.

  • [tex]v_i[/tex]= 235 m/s
  • t= 12 s
  • a= -10 m/s²

Substitute the values into the formula.

[tex]d= (235 \ m/s)(12 \ s) + \frac{1}{2} (-10 m/s^2)(12 \ s)^2[/tex]

Multiply the first 2 numbers in parentheses. The units of seconds cancel.

[tex]d=(235 \ m * 12 ) + \frac{1}{2} (-10 \ m/s^2)(12 \ s )^2[/tex]

[tex]d= (2820 \ m)+ \frac{1}{2} (-10 \ m/s^2)(12 \ s )^2[/tex]

Solve the exponent.

  • (12 s)²= 12 s * 12s = 144 s²

[tex]d=(2820 \ m)+ \frac{1}{2} (-10 \ m/s^2)(144 \ s^2)[/tex]

Multiply the other numbers in parentheses. The units of seconds squared cancel.

[tex]d=(2820 \ m)+ \frac{1}{2} (-10 \ m * 144 )[/tex]

[tex]d=(2820 \ m)+ \frac{1}{2} (-1440 \ m)[/tex]

Multiply by 1/2 or divide by 2.

[tex]d= 2820 \ m + (-720 \ m)[/tex]

Add.

[tex]d= 2100 \ m[/tex]

The firework traveled 2100 meters before exploding.

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