Q: A race car starting from rest accelerates uniformly at a rate of 4.90 meters per second what is the cars speed after it has traveled 200 meters?

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At the end of 7.508 seconds, the car has traveled 62 meters, and it's speed is 16.517 meters per second.

2 meters every second. bit pointless

If air resistance can be ignored, the distance in meters is 4.9t2. Note that 4.9 is half the numerical value of Earth's acceleration (9.8 meters per second square).

The average speed = 1/2 (initial speed + final speed) = 1/2 (15 + 25) = 1/2 (40) = 20 meters per second.The distance traveled = (average speed) x (time) = (20 x 7) = 140 meters.

5.0 meters every second.

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At the end of 7.508 seconds, the car has traveled 62 meters, and it's speed is 16.517 meters per second.

The formula for distance traveled with uniformly accelerated motion is d = 0.5 * a * t^2, where d is the distance, a is the acceleration, and t is the time. Plugging in the values: d = 0.5 * 40 * 8^2 = 0.5 * 40 * 64 = 1280 meters. Therefore, the car will travel 1280 meters in 8 seconds.

The distance traveled by the car can be calculated using the equation (s = \frac{1}{2}at^2), where (s) is the distance, (a) is the acceleration, and (t) is the time. Plugging in the values, we have (s = \frac{1}{2} \times 10 \times (4)^2 = 80) meters. Therefore, the car will travel 80 meters in 4 seconds.

2 meters every second. bit pointless

1249 + 2987 + 66245 = 70481

If air resistance can be ignored, the distance in meters is 4.9t2. Note that 4.9 is half the numerical value of Earth's acceleration (9.8 meters per second square).

60.912 meters in that time

The average speed = 1/2 (initial speed + final speed) = 1/2 (15 + 25) = 1/2 (40) = 20 meters per second.The distance traveled = (average speed) x (time) = (20 x 7) = 140 meters.

gravity

5.0 meters every second.

That's easy, if the car is initially traveling at 25 meters per second and gradually accelerates 3 meters per second for 6 seconds then the car is traveling at 43 meters per second.

23.6 meters per second.