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Margarita [4]
3 years ago
6

A dragster starts with zero velocity and completes a 404.5 m (0.2528 mile) run in 4.922 s. If the car had a constant acceleratio

n, what would be its acceleration and final velocity?
Physics
2 answers:
nikdorinn [45]3 years ago
6 0

Answer:

33.40 m/s² and 164.38 m/s²

Explanation:

From the equation of motion,

s = ut + 1/2at²................. Equation 1.

Where s = distance, u = initial velocity, a = acceleration, t = time.

Given: s = 404.5 m, t = 4.922 s, u = 0 m/s.

Substitute into equation 1

404.5 = 0(4.922) + 1/2(4.922²a)

404.5 = 12.11a

a= 404.5/12.11

a = 33.40 m/s²

Also using

v² = u² + 2as ................... Equation 2

Where v = final velocity.

Given: u = 0 m/s, a = 33.40 m/s². s = 404.5 m

Substitute into equation 2

v² = 0² + 2(33.40)(404.5)

v² = 27020.6

v = √(27020.6)

v = 164.38 m/s.

Hence  acceleration and the final velocity is 33.40 m/s² and 164.38 m/s respectively

KIM [24]3 years ago
5 0

Answer:

a. a=33.34ms⁻², V=164.4m/s

Explanation:

Since the dragster started with zero velocity, de determine the acceleration using of the equations of motion.

Below are the data given

Distance, s=404.5m,

time taken,t=4.922secs

Using the equation

S=ut+1/2at²

where u is the initial velocity and u=0

Making the acceleration the subject of the formula, we arrive at

a=2s/t²

a=(2*404.5)/4.922²

a=33.34ms⁻².

To determine the velocity, we use

V=u+at

V=0+33.34ms⁻² *4.922sec

V=164.4m/s

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a 2.0 kg block slides on the horizontal, frictionless surface until it counters a spring force constant with
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Complete question:

a 2.0 kg block slides on the horizontal, frictionless surface until it counters a spring with force constant of  955 N/m. The block comes to rest after compressing the spring a distance of 4.6 cm. Find the initial speed (in m/s) of the block.

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The initial speed of the block is 1.422 m/s

Explanation:

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