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A hover board is moving at 6 m/s . If it slows down at a rate of -3 m/s per second, how long is the board in the air before coming to rest? Round to 4 decimal places if necessary

Sagot :

Explanation

a uniformly accelerated motion is the one in which the acceleration of the particle throughout the motion is uniform,to fidn the time take we can use the formula:

[tex]\begin{gathered} t=\frac{v_f-v_1}{a} \\ where \\ v_f\text{ is the final velocity} \\ v_{i\text{ }}\text{ is the initail velocity} \\ a\text{ is the acceleration, so} \end{gathered}[/tex]

Step 1

a)Let

[tex]\begin{gathered} v_i=6\text{ }\frac{m}{s} \\ v_f=0(\text{ rest\rparen} \\ a=-3\frac{m}{s^2} \end{gathered}[/tex]

b) now, replace

[tex]\begin{gathered} t=\frac{v_{f}-v_{1}}{a} \\ t=\frac{0-(-6\frac{m}{s})}{3\frac{m}{s}} \\ t=2\text{ seconds} \end{gathered}[/tex]

therefore, the answer is 2 seconds

I

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