Hi, I need some assistance with the following questions:

Question: At the average rate of 15 m/s^2. How much time does it take for the complete increase in speed?
a. 0.25 s
b. 4.0 s
c. 0.0577 s
d. 17.3 s
e. 8.0 s

Question: A car traveling 20 m/s is able to stop in a distance d. Assuming the same braking force, what distance does this car require to stop when it is traveling twice as fast?
a. d
b. 2d
c. Squared 2d
d. 4d
e. Squared d

Solution Summary

This solution is comprised of the formulas needed for each of these physics-based problems and provides the algebra required to reach the final answer.

... Notice in the question that the distances are divided into equal halves. ... Here the total distance is divided in to two halves. ...Solving we get as. ...

... in a vacuum, a is the acceleration of gravity, g. If we record the time required for an object to fall a distance s in a time t, we can solve for g. Using the ...

... Solution: Let the time taken for walking be t then time taken for jogging will be (1/2 - t). Distance of walking = speed x time = 4t Distance of ...

... Use the d = rt formula, and solve for t (t = d/r) to find the times for each train. ... Train: distance = x miles rate = 40 miles/hour time = x/40 hours. Bus: ...

... Find the distance travelled by the object during the first 2.10 s ... of a polynomial for time (t) that defines the velocity at time t. A solution that involves ...

...distance / time = 560/(x+1) Speed in the second portion = distance / time = 480/x. Since speed is same on both portion, we have 560 480 = x +1 x. Solving for ...

... Use this formula to solve the problem ... the direction of the current, the overall speed is (3+x). Therefore, the time taken to cover the 3 mile distance in the ...

... ball and ground is 0.75, determine the distance d when ... t when the ball struck the ground for the second time. Please show all work and solve it just for theta ...