Math/Physics

Question

Suppose a car merges into freeway traffic on a 200 long ramp. If its initial velocity is 10 m/s and it accelerates at 2m/s², how long does it take to travel 200 m up the ramp?

Please provide a detailed explainiation/work.

 

For the following question, I completed it and I just wanted to see if the answer was correct. 

A jogger runs down a straight stretch of a road with an average velocity of 4m/s for 2.0 minutes. What is the final position taking the initial position to be 0.

MY ANSWER:

v=x-x_{0/t    ; v is} average velocity; x is final position; x₀ is the initial position; t is time..

4=x-0/2

(2) 4=x-0

8=x-0

0+8=x

8=x

The final position of the jogger is 8.

Answer 1

Rebecca, you must take the units into account. 4m/s is equivalent to 4*60 m/min =240 m/min. So after 2 mins the jogger has travelled 480 m.

 

For the other problem, we can use the formula x = v_ot + 1/2at², where x = distance, a = accel., and t = time,                            so 200 = 10t +1/2(2)t² or 200 = 10t + t²

Rearranging, we get: t² + 10t - 200 = 0

This factors into (t + 20)(t - 10) = 0

So t = 10 (or -20, which is not a sensible solution)

So it takes 10 sec.