Question

An alligator swims to the left with a constant velocity of 5 \,\dfrac{\text{m}}{\text s}5
s
m
​
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
​
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
​
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
​
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
​
35, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction

Answer 1

Answer:

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

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

$v_f = v_i + at$

$35 = 5 + 3 t$

$t = 10 s$