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adelina 88 [10]
3 years ago
10

A car is traveling with a constant speed when the driver suddenly applies the brakes, giving the 14) car a deceleration of 3.50

m/s2. if the car comes to a stop in a distance of 30.0 m, what was the car's original speed?
Physics
1 answer:
sergij07 [2.7K]3 years ago
7 0
To be able to determine the original speed of the car, we use kinematic equations to relate the acceleration, distance and the original speed of the car moving. 

First, we manipulate the one of the kinematic equations
 
v^2 = v0^2 + 2 (a) (x)  where v = 0 since the car stopped

Writing the equation in such a way that the initial velocity or v0 is written on one side of the equation,

<span>we get v0 = sqrt (2(a)(x))

Substituting the known values,

v0 = sqrt(2(3.50)(30.0))
v0 = 14.49 m/s 
</span>
Therefore, before stopping the car the original speed of the car would be 14.49 m/s
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Answer:

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y=65t-16t^{2}

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y_{t=2}=65(2)-16(2)^{2} =66ft

y_{t=2.5}=65(2.5)-16(2.5)^{2} =62.5ft

v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s

--

y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft

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b) To find the instantaneous velocity we need to derivate the equation

v=\frac{df}{dt}=65-32t

v=65-32(2)=1ft/s

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