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Deffense [45]
3 years ago
12

The diagram shows a position-time graph What is the displacement of the object

Physics
1 answer:
algol133 years ago
5 0

The object is moving, so at different times, it has different displacement.  I'm guessing that you probably want to know the displacement at the end of the time on the graph ... 5 seconds.

Displacement is the distance and the direction FROM (the position at the  beginning) TO (the position at the end).

At the beginning ... time=0 ... the position is 1 meter.

At the end ... time=5 ... the position is zero.

The distance FROM the beginning TO the end is (zero - 1m) .  That's  <em>-1m </em>.


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Human reaction times are worsened by alcohol. How much further (in feet) would a drunk driver's car travel before he hits the br
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Answer:

A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes

Explanation:

Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:

2as = Vf² - Vi²

s = (Vf² - Vi²)/2a

where,  

Vf = Final Velocity of Car = 0 mi/h

Vi = Initial Velocity of Car = 50 mi/h

a = deceleration of car  

s = distance covered

Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.

So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:

s = vt

FOR SOBER DRIVER:

v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 0.33 s

s = s₁

Therefore,

s₁ = (73.33 ft/s)(0.33 s)

s₁ = 24.2 ft

FOR DRUNK DRIVER:

v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 1 s

s = s₂

Therefore,

s₂ = (73.33 ft/s)(1 s)

s₂ = 73.33 ft

Now, the distance traveled by drunk driver's car further than sober driver's car is given by:

ΔS = s₂ - s₁

ΔS = 73.33 ft - 24.2 ft

<u>ΔS = 49.13 ft</u>

6 0
3 years ago
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