The acceleration of the bicyclist is equal to [tex]a=2\frac{m}{s^{2} }[/tex] to the left.
Why?
I'm assuming that you forgot to write the question of the problem, so, I'm completing it. Also, I noticed that there is a mistake, you wrote "a steady gust of wind that lasts 3.5 m/s", it should be "a steady gust of wind that lasts 3.5 s", since we are talking about time here.
The entire and corrected statement would be:
"A bicyclist is riding to the left with a velocity of 14m/s after a steady gust of wind that lasts 3.5s the bicyclist is moving to the left with a velocity of 21m/s. Assuming the acceleration is constant, what is the acceleration of the bicyclist?
Now, we can see that the velocity of the bicyclist increased after the steady gust, knowing that, we can assume that the steady gust was following the same direction of the cyclist.
We can solve the problem using the following formula:
[tex]V_{f}=V_{o}+a*t[/tex]
From the statement we know that:
[tex]V_{f}=V_{2}=21\frac{m}{s}\\\\V_{o}=V_{1}=14\frac{m}{s}\\\\time=t=3.5s[/tex]
So, substituting and isolating the acceleration, we have:
[tex]21\frac{m}{s}=14\frac{m}{s}+a*(3.5s)[/tex]
[tex]21\frac{m}{s}-14\frac{m}{s}=a*(3.5s)[/tex]
[tex]7\frac{m}{s}=a*(3.5s)[/tex]
[tex]a=\frac{7\frac{m}{s} }{3.5s}=\frac{7m}{3.5s*s}=2\frac{m}{s^{2} }[/tex]
Hence, we know that the acceleration of the bicyclist is equal to:
[tex]a=2\frac{m}{s^{2} }[/tex] to the left.
Have a nice day!
