Question

Write your question here (Keep it simple and clear to get the best answer) A rececar starting from rest accelerates at a constantrate of 5.00 m/s²  1,what is the velocity of the car after it has traveled ft 2,How much time has elopsed 3,calculate the average velocity in two different ways

Answer 1

I assume by 1.00, you mean 1.00 second.

Let’s assume there’s no account for friction in this assessment.

The formula is:

Distance=(1/2)(acceleration)(time)^2

So let’s solve it.

S = (1/2)(5 meters)(1 seconds)^2

S = 2.5 meters

Now you also expressed 102 ft as a guess, the acceleration required to go 102 feet is found as follows:

102ft =(1/2) (a) (1sec)^2

102ft/(1/2)(1sec)^2 = a

51 ft/sec^2

Also to find time at that given acceleration rate, to reach 102 feet. We first must convert feet to meters:

102feet(0.3048meter/1foot)

The feet cancel and we’re left with 31.0896 meters, for calculation accuracy we’ll keep it this way.

Let s be distance traveled

t = (sqrt(2(a)(s)))/a

t = (sqrt(2(5)(31.0896)))/5

t = 3.526449 seconds

Hope I answered all your question