Help me solve the following question.

x axis is x = 4.00 - 7.00t2, with x in meters and t in seconds. (a) At what time and (b) where does the particle (momentarily) s

andrezito [222]

Answer:

(a) 0 s

(b) 4.00 m

(c) -0.76 s

(d) +0.76 s

Step-by-step explanation:

$x=4.00-7.00t^2$

It stops momentarily when the velocity, $v$ is 0. $v$ is the derivate of $x$ .

(a) $v=\frac{dx}{dt}=-14.00t$

Setting this to 0,

$-14.00t=0$

$t=0$

(b) Substitute this value for $t$ in $x$ to get its position.

$x=4.00-7.00\times0^2=4.00$ m

It passes the origin when $x=0$

$4.00-7.00t^2=0$

$7.00t^2=4$

$t^2=\frac{4}{7}$

$t=\pm\sqrt{\frac{4}{7}}$

(c) The negative time is $t=-\sqrt{\frac{4}{7}} =-0.76 s$

(d) The positive time is $t=+\sqrt{\frac{4}{7}} =+0.76 s$

8 0

4 years ago

Find E in DEF<br> D=5, E=3x+14

vova2212 [387]

Answer:

well, I think it is B 7°

as X =7

because 5x=3x+14

so 5x7= 35

3x7=21 + 14=35

6 0

3 years ago

Can someone help me !!!

nordsb [41]

RM=10 PM is symmetrical to it

7 0

4 years ago

WORK THIS CHALLENGE EXERCISE. A is known to be 6,500 feet above sea level; AB = 600 feet. The angle at A looking up at P is 20°.

adell [148]

Answer:

6955ft

T

Step-by-step explanation:

Hope that helps

6 0

3 years ago

Read 2 more answers

The pre-image of pentagon PENTA is rotated 90° clockwise about the origin to create the image of pentagon P'ENTA. Which statemen

belka [17]

Yeah dude ima be honesti think it's 145 but don't get mad if it's not right , your welcome

5 0

3 years ago