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Lyrx [107]
3 years ago
5

A bicyclist is riding to the left with a velocity of 14 \,\dfrac{\text m}{\text s}14 s m ​ 14, start fraction, start text, m, en

d text, divided by, start text, s, end text, end fraction. After a steady gust of wind that lasts 3.5\,\text s3.5s3, point, 5, start text, s, end text, the bicyclist is moving to the left with a velocity of 21\,\dfrac{\text m}{\text s}21 s m ​ 21, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction. Assuming the acceleration is constant, what is the acceleration of the bicyclist?
Physics
1 answer:
Gnesinka [82]3 years ago
5 0

Answer:

-2.0 m/s²

Explanation:Acceleration is the rate of change of velocity.

\begin{aligned}a&=\dfrac{\text{Change in velocity}}{\text{Change in time}}\\ \\ &=\dfrac{v_f-v_i}{\Delta t} \end{aligned}

a

​

 

=

Change in time

Change in velocity

​

=

Δt

v

f

​

−v

i

​

​

​

Hint #22 / 3

We can calculate the bicyclist's acceleration from the final velocity v_fv

f

​

v, start subscript, f, end subscript, initial velocity v_iv

i

​

v, start subscript, i, end subscript, and time interval \Delta tΔtdelta, t.

\begin{aligned}a&=\dfrac{v_f-v_i}{\Delta t}\\ \\ &=\dfrac{-21\,\dfrac{\text m}{\text s}-(-14\,\dfrac{\text m}{\text s})}{3.5\,\text s}\\ \\ &=-2.0\,\dfrac{\text m}{\text s^2}\end{aligned}

a

​

 

=

Δt

v

f

​

−v

i

​

​

=

3.5s

−21

s

m

​

−(−14

s

m

​

)

​

=−2.0

s

2

m

​

​

Hint #33 / 3

The acceleration of the bicyclist is -2.0\,\dfrac{\text m}{\text s^2}−2.0

s

2

m

​

minus, 2, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction.

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