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JulijaS [17]
3 years ago
12

Dragsters can actually reach a top speed of 145.0 m/s in only 4.45 s. (a) Calculate the average acceleration for such a dragster

. (b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402.0 m (a quarter mile) without using any information on time. (c) Why is the final velocity greater than that used to find the average acceleration?
Physics
1 answer:
astraxan [27]3 years ago
7 0

Answer:

a) 32.58 m/s²

b) 161.84 m/s

Explanation:

Initial velocity = u = 0

Final velocity = v = 145 m/s

Time taken = t = 4.45 s

s = Displacement of dragster = 402 m

a = Acceleration

v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{145-0}{4.45}\\\Rightarrow a=32.58\ m/s^2

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as-u^2}\\\Rightarrow v=\sqrt{2\times 32.58\times 402-0^2}\\\Rightarrow v=161.84\ m/s

The final velocity is greater than the velocity used to find the average acceleration due to the gear changes. The first gear in a dragster has the most amount of toque which means the acceleration will be maximum. The final gears have less torque which means the acceleration is lower here. The final gears have less acceleration but can spin faster which makes the dragster able to reach higher speeds but slowly.

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now,

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1600 rpm =  = 570 \times \dfrac{2\pi}{60}

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3 years ago
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Initial velocity vector vA has a magnitude of 3.00 meters per second and points 20.0o north of east, while final velocity vector
garri49 [273]

Answer:

5.2\ \text{m/s}

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Explanation:

v_a = 3 m/s

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v_{ax}=v_a\cos \theta\\\Rightarrow v_{ax}=3\times \cos 20^{\circ}\\\Rightarrow v_{ax}=2.82\ \text{m/s}

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x and y component of v_b

v_{bx}=v_b\cos \theta\\\Rightarrow v_{bx}=6\times \cos 320^{\circ}\\\Rightarrow v_{bx}=4.6\ \text{m/s}

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Magnitude

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Direction

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The magnitude of the change in velocity vector is 5.2\ \text{m/s} and the direction is 70^{\circ} south of east.

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