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Tcecarenko [31]
2 years ago
12

If a car can go from 0 to 60 km/h in 8.0 seconds, what would be its final speed after 5.0 seconds if its starting speed were 50

km/h?
Physics
1 answer:
dmitriy555 [2]2 years ago
3 0

Answer:

This question assumes that the car accelerates at the same rate as when it went from 0 to 60km/h

24.29m/s or 87.4km/h

Explanation:

Let's find the acceleration of the car:

let vi=0, vf=60km/h (16.67m/s), Δt = 8.0s

a = (vf-vi)/Δt

a = (16.67m/s-0)/8.0

a = 2.08m/s^2

Now we can use this acceleration to find vf in the second part:

50km/h is 13.89m/s

a = (vf-vi)Δt

vf = aΔt + vi

vf = 2.08m/s^2*5.0+13.89m/s

vf = 24.29m/s (87.4km/h)

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A bike first accelerates from 0.0m/s to 4.5m/s in 4.5 s, the continues at this constant speed for another 6.0 s. What is the tot
Alex_Xolod [135]

Answer:

37.125 m

Explanation:

Using the equation of motion

s=ut+0.5at^{2} where s is distance, u is initial velocity, t is time and a is acceleration

<u>Distance during acceleration</u>

Acceleration, a=\frac {V_{final}-V_{initial}}{t} where V_{final} is final velocity and V_{initial} is initial velocity.

Substituting 0.0 m/s for initial velocity and 4.5 m/s for final velocity, acceleration will be

a=\frac {4.5 m/s-0 m/s}{4.5 s}=1 m/s^{2}

Then substituting u for 0 m/s, t for 4.5 s and a for 1 m/s^{2} into the equation of motion

s=0*4.5+ 0.5*1*4.5^{2}=0+10.125 =10.125 m

<u>Distance at a constant speed</u>

At a constant speed, there's no acceleration and since speed=distance/time then distance is speed*time

Distance=4.5 m/s*6 s=27 m

<u>Total distance</u>

Total=27+10.125=37.125 m

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3 years ago
A 1.00 kg object is attached to a horizontal spring. the spring is initially stretched by 0.500 m, and the object is released fr
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The  spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
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v_{max} = A \omega = (0.500 m)(31.4 rad/s)= 15.7 m/s
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