The minimum stopping distance when the car is moving at 32.0 m/s is 348.3 m.
<h3>
Acceleration of the car </h3>
The acceleration of the car before stopping at the given distance is calculated as follows;
v² = u² + 2as
when the car stops, v = 0
0 = u² + 2as
0 = 15² + 2(76.5)a
0 = 225 + 153a
-a = 225/153
a = - 1.47 m/s²
<h3>Distance traveled when the speed is 32 m/s</h3>
If the same force is applied, then acceleration is constant.
v² = u² + 2as
0 = 32² + 2(-1.47)s
2.94s = 1024
s = 348.3 m
Learn more about distance here: brainly.com/question/4931057
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The definition of volocity is <span>the speed of something in a given direction.</span>
we know
T=2.pie sqr root(l/g)
l= (sqrof 7/22)*9.8
L= .99 or approx 1m
Without what values please be more specific
Answer:
safe speed for the larger radius track u= √2 v
Explanation:
The sum of the forces on either side is the same, the only difference is the radius of curvature and speed.
Also given that r_1= smaller radius
r_2= larger radius curve
r_2= 2r_1..............i
let u be the speed of larger radius curve
now, 
................ii
form i and ii we can write
⇒u= √2 v
therefore, safe speed for the larger radius track u= √2 v
