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stepladder [879]
3 years ago
9

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

er B starts from rest and accelerates at 2.0 m/s^2 for the first 30 m and then runs at constant speed. Runner A befins running as soon as the race begins but B first takes a nap to rest up. What is the longest nap that B can take and still not lose the race?
Physics
1 answer:
kifflom [539]3 years ago
4 0

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

s=ut+\dfrac{1}{2}at_{A}^2

Put the value in the equation

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

t_{A}=\sqrt{\dfrac{30\times2}{1.6}}

t_{A}=6.12\ sec

We need to calculate the time for B

Using equation of motion

s=ut+\dfrac{1}{2}at_{B}^2

Put the value in the equation

30=0+\dfrac{1}{2}\times2.0\times t_{B}^2

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

t_{B}=5.48\ sec

We need to calculate the time for longest nap

Using formula for difference of time

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

t'=6.12-5.48

t'=0.64\ s

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

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Another way to think of it is that biker A gains 1.7 meters on B every second (6.4-4.7=1.5), so the time it'll take for him to gain 34 meters is 34/1.7=20 seconds. In that time, biker B travels 4.7*20=94 meters
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coldgirl [10]

Answer:

A) t = 0.55 s

B) x = 24.8 m

Explanation:

A) We can find the time at which the ball will be in the air using the following equation:

y_{f} = y_{0} + v_{0y}t - \frac{1}{2}gt^{2}    

Where:

y_{f} is the final height= 0  

y_{0} is the initial height= 1.5 m

v_{0y} is the component of the initial speed in the vertical direction = 0 m/s        

t: is the time =?      

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0 = 1.5 m - \frac{1}{2}9.81 m/s^{2}t^{2}

By solving the above equation for t we have:

t = \sqrt{\frac{2*1.5 m}{9.81 m/s^{2}}} = 0.55 s  

Hence, the ball will stay 0.55 seconds in the air.

                             

B) We can find the distance traveled by the ball as follows:

x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2}

Where:  

a: is the acceleration in the horizontal direction = 0  

x_{f} is the final position =?  

x_{0} is the initial position = 0      

v_{0x} is the component of the initial speed in the horizontal direction = 45 m/s                                                                                            

x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2}

x_{f} = 0 + 45 m/s*0.55 s + 0 = 24.8 m

Therefore, the ball will travel 24.8 meters.

I hope it helps you!

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