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tamaranim1 [39]

3 years ago

13

the school runningback, gets the ball at the 50 yard line. He gains 40 yards toward the goal line and then gets pushed back 5 ya

rds and tackled.

2 answers:

blondinia [14]3 years ago

8 0

Answer: he is at the 15 yard line

Step-by-step explanation: once he gains 40 yards he is at the 10 yard line then once he goes back 5 he is at the 15 yard line which is in between the 20 and ten

VikaD [51]3 years ago

4 0

Answer:

He would be at the 15 yard line.

Step-by-step explanation:

The 40 yard gain would bring him to the 10 yard line (he caught it at the 50 yard line, so 50 - 40 = 10), but the 5 yards lost due to the tackle and push back would bring him to the 15 yard line (10 + 5 = 15).

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The position of an object moving along an x axis is given by x = 3.24 t - 4.20 t2 + 1.07 t3, where x is in meters and t in secon

Pavel [41]

Answer and explanation:

Given : The position of an object moving along an x axis is given by $x=3.24t-4.20t^2+1.07t^3$ where x is in meters and t in seconds.

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a) At t= 1 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(1)=3.24(1)-4.20(1)^2+1.07(1)^3$

$x(1)=3.24-4.20+1.07$

$x(1)=0.11$

b) At t= 2 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(2)=3.24(2)-4.20(2)^2+1.07(2)^3$

$x(2)=6.48-16.8+8.56$

$x(2)=-1.76$

c) At t= 3 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(3)=3.24(3)-4.20(3)^2+1.07(3)^3$

$x(3)=9.72-37.8+28.89$

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d) At t= 4 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(4)=3.24(4)-4.20(4)^2+1.07(4)^3$

$x(4)=12.96-67.2+68.48$

$x(4)=14.24$

(e) What is the object's displacement between t = 0 and t = 4 s?

At t=0, x(0)=0

At t=4, x(4)=14.24

The displacement is given by,

$\triangle x=x(4)-x(0)$

$\triangle x=14.24-0$

$\triangle x=14.24$

(f) What is its average velocity from t = 2 s to t = 4 s?

At t=2, x(2)=-1.76

At t=4, x(4)=14.24

The average velocity is given by,

$\triangle x=x(4)-x(2)$

$\triangle x=14.24-(-1.76)$

$\triangle x=14.24+1.76$

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2 years ago

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