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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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Consider a block on a spring oscillating on a frictionless surface. The amplitude of the oscillation is 11 cm, and the speed of
IRISSAK [1]

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Explanation:

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2 years ago
In a 350-m race, runner A starts from rest and accelerates at 1.6 m/s^2 for the first 30 m and then runs at constant speed. Runn
kifflom [539]

Answer:

B can take 0.64 sec for the longest nap .

Explanation:

Given that,

Total distance = 350 m

Acceleration of A = 1.6 m/s²

Distance = 30 m

Acceleration of B = 2.0 m/s²

We need to calculate the time for A

Using equation of motion

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Put the value in the equation

30=0+\dfrac{1}{2}\times1.6\times t_{A}^2

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We need to calculate the time for B

Using equation of motion

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Put the value in the equation

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t_{B}=\sqrt{\dfrac{30\times2}{2.0}}

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Using formula for difference of time

t'=t_{A}-t_{B}

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t'=0.64\ s

Hence, B can take 0.64 sec for the longest nap .

4 0
3 years ago
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