Physics, 21.01.2021 23:00 rakanmadi87

At a sports car rally, a car starting from rest accelerates uniformly at a rate of 5 m/s/s over a straight-line distance of 291 m. How long (in seconds) did it take the car to travel the 291 m

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Answer from: 3345

The time taken by the car to travel 291 m is calculated using the kinematics equation and it is obtained to be 10.8 seconds.

Motion in a Straight Line

Here, it is given that;

The acceleration of the car is,

$a = 5\,m/s^2$

The distance covered by the car is,

$s = 291\,m$

Also, the car starts from rest, that means;

$u = 0\,m/s$

We can use the second kinematics equation to find the time taken by the car to travel the given distance.

$s = ut + \frac{1}{2}\,at^2\\291\,m = (0\,m/s \times t)+ \frac{1}{2} (5\,m/s^2\times t^2)\\\\\\291\,m = (2.5\,m/s^2\times t^2)$

implies

$t = \sqrt{\frac{291\,m}{2.5\,m/s^2}}=10.78\,s\approx10.8\,s$

Learn more about 'motion in a straight line' here: link

Answer from: robert7248

10.8 s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Acceleration (a) = 5 m/s/s

Distance travelled (s) = 291 m

Time (t) taken =?

We can calculate the time taken for the car to cover the distance as follow:

s = ut + ½at²

291 = 0 × t + ½ × 5 × t²

291 = 0 + 2.5 × t²

291 = 2.5 × t²

Divide both side by 2.5

t² = 291 / 2.5

t² = 116.4

Take the square root of both side

t = √116.4

t = 10.8 s

Thus, it will take the car 10.8 s to cover the distance.

Answer from: Quest

answer and explanation;

composition refers to both the types of minerals within a rock and the overall chemical makeup of the rock while texture refers to the size and arrangement of the minerals or grains that make up a rock. both composition and texture can be used to classify rocks, such as igneous rocks. texture describes how the rock became solid or crystallized from liquid, melted rock. composition of a rock describes the minerals that make up a particular rock.

Answer from: Quest

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

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