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A sports car accelerates uniformly from rest to a speed of 87 mi/hr in 8s. Determine: a.The acceleration of the car
b.The distance the car travels in the first 8 s
c.The velocity of the car after the first 10 s

Respuesta :

Answer:

Part a)

[tex]a = 4.86 m/s^2[/tex]

Part b)

[tex]d = 155.52 m[/tex]

Part c)

[tex]v_f = 48.6 m/s[/tex]

Explanation:

As we know that car start from rest and reach to final speed of 87 mph

so we have

[tex]v_f = 87 mph = 38.88 m/s[/tex]

now we have

Part a)

acceleration is rate of change in velocity

[tex]a = \frac{v_f - v_i}{t}[/tex]

[tex]a = \frac{38.88 - 0}{8}[/tex]

[tex]a = 4.86 m/s^2[/tex]

Part b)

distance moved by car with uniform acceleration is given as

[tex]d = \frac{v_f + v_i}{2} t[/tex]

[tex]d = \frac{38.88 + 0}{2} 8[/tex]

[tex]d = 155.52 m[/tex]

Part c)

As we know that the car start from rest

so final speed after t = 10 s

[tex]v_f = v_i + at[/tex]

[tex]v_f = 0 + (4.86)10[/tex]

[tex]v_f = 48.6 m/s[/tex]

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