Answer:
1) acceleration = 1.3 [tex]\frac{m}{s^2}[/tex]
2) acceleration = 26.1 [tex]\frac{m}{s^2}[/tex]
3) acceleration = 2.5 [tex]\frac{m}{s^2}[/tex]
4) Final velocity = 13 m/s
5) Time = 0.7 seconds
Explanation:
1) First question:
Acceleration can be calculated based on the formula:
[tex]a=\frac{\Delta v}{\Delta t} =\frac{15-2.9}{9} \approx 1.3444\,\,\frac{m}{s^2}[/tex]
which rounded to one decimal is: 1.3 m/s^2
2) Second question:
Use the formula that relates final and initial velocity with acceleration and distance covered:
[tex]v_f^2-v_i^2=2\,*\,a\,* \,distance\\15^2-4^2=2\,a\,(4)\\209=8 \, a\\a = \frac{209}{8} \,\frac{m}{s^2} \\a=26.125\,\,\frac{m}{s^2}[/tex]
which rounded to one decimal gives:
acceleration = 26.1 [tex]\frac{m}{s^2}[/tex]
3) Third question :
The marble rolls down 5 meters in 2 seconds, so we use the distance covered formula under constantly accelerated motion:
[tex]distance=v_i\,t+\frac{1}{2} a\,t^2\\5 = 0 +\frac{1}{2} a\,(2)^2\\5 = 2 a\\a = 2.5 \,\frac{m}{s^2}[/tex]
4) Fourth question:
Use the formula
[tex]v_f^2-v_i^2=2\,*\,a\,* \,distance\\v_f^2=7^2+2\,(6)\,(10)\\v_f^2=169\\v_f=\sqrt{169} \\v_f=13\,\frac{m}{s}[/tex]
5) Fifth question:
First calculate the acceleration :
[tex]v_f^2-v_i^2=2\,*\,a\, distance\\18.8^2-2.21^2=2\,a\,(60)\\349.03=120\,a\\a=2.908\,\frac{m}{s^2}[/tex]
Now we can find the time it takes the car to reach a speed of 4 m/s via the formula:
[tex]v_f-v_i=a\,t\\4-2.1=2.908\,t\\t = 1.9 / 2.908 \,\,s\\t = 0.65337 \,\,s[/tex]
which rounded to one decimal gives:
time = 0.7 seconds
