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2 years ago

a car starts from the rest and accelerates at 9.54m/s for 6.5 seconds. what is the distance covered by the car

Physics

1 answer:

uysha [10]2 years ago

3 0

Answer:

= 201.53 meters

Explanation:

A car started from rest and accelerated at 9.54 m/s^2 for 6.5 seconds. How much distance was covered by the car?

Use the formula d = $\frac{at^{2} }{2} ,$

where d is the distance, t is the time and "a" is the acceleration.
$d=\frac{9*54*6*5^{2} }{2} = 201.53 m$

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quester [9]

The image is missing, so i have attached it;

Answer:

A) V_rel = [-(2.711)i - (0.2588)j] m/s

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Explanation:

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Angular velocity; ω = 1 deg/s = 1° × π/180 rad = 0.01745 rad/s

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0 = (α*r_ba) + (ω(ω•r_ba)) + (2ω*v_rel) + a_rel

a_rel = - (α*r_ba) - (ω(ω•r_ba)) - (2ω*v_rel)

Plugging in the relevant values with their respective position vectors, we have;

a_rel = - (-0.008727k * -100j) - (0.01745k(0.01745k * -100j)) - (2*0.01745k * [-(2.711)i - (0.2588)j])

a_rel = 0.8727i - (0.01745² x 100)j + 0.0946j - 0.009i

Note that; k x j = - i and k x i = j

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a_rel = 0.8637i + 0.0642j m/s²

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a car starts from the rest and accelerates at 9.54m/s for 6.5 seconds. what is the distance covered by the car​

a car starts from the rest and accelerates at 9.54m/s for 6.5 seconds. what is the distance covered by the car