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Iteru [2.4K]
3 years ago
14

A moving cyclist sees traffic up ahead and begins slowing down with constant acceleration. After slowing down for 0.5\,\text{s}0

.5s0, point, 5, start text, s, end text, the cyclist has travelled forward 2\,\text m2m2, start text, m, end text and has a speed of 10\,\dfrac{\text{km}}{\text{h}}10 h km ​ 10, start fraction, start text, k, m, end text, divided by, start text, h, end text, end fraction. We want to find the acceleration of the cyclist while slowing down to 10\,\dfrac{\text{km}}{\text{h}}10 h km ​ 10, start fraction, start text, k, m, end text, divided by, start text, h, end text, end fraction.
Physics
1 answer:
Helen [10]3 years ago
8 0

Answer:

Δx=vt−21​at2

Explanation:

We can figure out which kinematic formula to use by choosing the formula that includes the known variables, plus the target unknown.

In this problem, the target unknown is the acceleration as the cyclist slows down.

Assuming the initial direction of travel is the positive direction, our known variables are

t=0.5s

v=10km/h

Δx=2m

Since we don't know the initial velocity (Vo), and we are not asked to find (Vo)  we could use the kinematic formula that is missing (Vo) to solve for the target unknown, A

Δx=vt−21​at

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A baseball pitcher throws the ball towards the batter at 90 mph. His bat connects with the ball for a line drive, after which th
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Answer:

F=-18412.9N, where the minus indicates the direction is opposite to that of the throw.

Explanation:

a)

Since MKS stands for meter-kilogram-second and we know that:

1\ hour = 3600\ seconds

1\ mile = 1600\ meters

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We can write that:

\frac{1\ hour}{3600\ seconds}=1

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These are conversion factors, equal to 1, so multiplying our results by them won't change their value, only their units.

So we have that:

90 mph=90 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=40m/s

110 mph=110 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=48.89m/s

145 g=145 g(\frac{1kg}{1000g})=0.145kg

b)

Newton's 2nd Law tells us that F=ma, and the definition of acceleration is a=\frac{\Delta v}{\Delta t}, so we have:

F=m\frac{\Delta v}{\Delta t}=m\frac{v_f-v_i}{t}

Taking the throw direction as the positive one, for our values we have:

F=m\frac{v_f-v_i}{t}=(0.145kg)\frac{(-48.89m/s)-(+40m/s)}{0.0007s}=-18412.9N

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