Question

An alligator swims to the left with a constant velocity of 5 \,\dfrac{\text{m}}{\text s}5

s
m
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5, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction. When the alligator sees a bird straight ahead, the alligator speeds up with a constant acceleration of 3 \,\dfrac{\text {m}}{\text s^2}3
s
2

m
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3, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction leftward until it reaches a final velocity of 35 \,\dfrac{\text {m}}{\text s}35
s
m
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35, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction leftward.
How many seconds does it take the alligator to speed up from 5 \,\dfrac{\text {m}}{\text s}5
s
m
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5, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction to 35 \,\dfrac{\text {m}}{\text s}35
s
m
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35, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction

Answer 1

The alligator will take t = 10 s to reach the final speed of 35 m/s

Step-by-step explanation:

As we know that the initial speed of the alligator is 5 m/s

then it accelerate by given acceleration to reach the final speed of 35 m/s

so we will have

`v_i = 5 m/s`

`v_f = 35 m/s`

`a = 3m/s^2`

now we have

`v_f = v_i + at`

`35 = 5 + 3 t`

`t = 10 s`