Before the driver applies the brakes ( with the reaction time ):
d 1 = v0 · t = 20 m/s · 0.53 s = 10.6 m
After that:
v = v0 - a · t1
0 = 20 m/s - 7 · t1
7 · t1 = 20
t1 = 2.86 s
d 2 = v 0 · t1 - a · t1² / 2
d 2 = 20 m/s · 2.86 s - 7 m/s² · (2.86 s)²/2 = 57.2 m - 28.6 m = 28.6 m
d = d 1 + d 2 = 10.6 m + 28.6 m = 39.2 m
Answer: the stopping distance of a car is 39.2 m.
<u>Q</u><u>uest</u><u>ion</u><u>:</u>
To Simplify:
118 {121÷(11×11)-(-4)-(3-7)}
<u>Solu</u><u>tion</u>:
↠118 {121÷121+4-(-4)}
↠118 {1+4+4}
↠118 {5+4}
↠118{9}
↠118×9
↠1062
▬▬▬▬▬▬▬▬▬▬▬▬
The upper Solution is done by applying BODMAS
<u>Abou</u><u>t</u><u> </u><u>BODMAS</u><u>:</u>
B→ Bracket
O→ Of
D→ Division
M→ Multiplication
A→ Addition
S→ Subtraction
Answer:
3,498.921
Step-by-step explanation:
Answer:
wait it is 325 of 3.25
Step-by-step explanation:
I can't tell if your asking for just one number, which would be 3.25...
A = 11300 cm<span>2
</span>
Hope this helps!