Physics, 18.09.2019 19:00 black4sureg

3. a bike rider accelerates uniformly at 2.0 m/s2 for 10.0
s. if the rider starts from rest, calculate the distance
traveled in the fourth second.
(i. e. between t = 3 s and t = 4 s).

Answers: 1

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Answer from: babyskitt

16 m

Explanation:

Since this situation is about constanta acceleration, we can use the following equations:

$V=V_{o} +a.t$ (1)

$V^{2}={V_{o}}^{2} +2ad$ (2)

Where:

is the bike rider's final velocity

$V_{o}=0$ is the bike rider's initial velocity

t=4 s is the time at which we need to find the bike rider's distance

$a=2 \frac{m}{s^{2}}$ is the constant acceleration

is the bike rider's traveled distance

From (1) we can find :

V=0 +a.t (3)

$V=(2 \frac{m}{s^{2}})(4 s)$ (4)

$V=8\frac{m}{s}$ (5)

Now, we can substitute (5) in (2) and find the traveled distance :

$V^{2}=0 +2ad$ (6)

$d=\frac{V^{2}}{2a}$ (7)

Finally:

d=16 m

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3. a bike rider accelerates uniformly at 2.0 m/s2 for 10.0
s. if the rider starts from rest, c...

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