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A car driving at an initial speed of 10.0 m/s accelerates on a straight road at 3.00 m/s^2. a) what is the speed of the car after one quarter of a mile? (1 mile =1.609km). B) the driver in part A slams on the brakes after reaching the quarter mile. If the car can decelerate at a rate of 4.50 m/s^2, what is the stopping distance of the car?

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Answer:

The speed of the car and the stopping distance are 50.13 m/s and 279.22 m.

Step-by-step explanation:

Given that,

Initial speed = 10.0 m/s

Acceleration = 3.00 m/s^2

Distance
d = (1)/(4)*1.609


d = 0.40225\ km=0.40225*10^(3)\ m

We need to calculate the speed of the car,

Using equation of motion


v^2-u^2=2as

Where, u = initial velocity

v = final velocity

a = acceleration

Put the value in the equation


v^2=(10.0)^2+2*3.00*0.40225*10^(3)


v^2=2513.5


v=50.13\ m/s

(B). We need to calculate the stopping distance of the car,

Using equation of motion again


v^2=u^2+2as

Here,initial velocity = 50.13 m/s

Final velocity = 0

Acceleration = -4.50 m/s²

Put the value in the equation


0=(50.13)^2-2*4.50* s


s=((50.13)^2)/(2*4.50)


s=279.22\ m

Hence, The speed of the car and the stopping distance are 50.13 m/s and 279.22 m.

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