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LenaWriter [7]
2 years ago
5

A car slows down at -5.00 m/s2 until it comes to a stop after traveling 15.0 m. How much time did it take to stop.

Mathematics
1 answer:
professor190 [17]2 years ago
4 0

Answer:

2.45 seconds to the nearest hundredth.

Step-by-step explanation:

Acceleration a = -5, distance s = 15, time  = ?.

Final  velocity = v =  0.

We need to use the equations of motion under constant acceleration.

v^2 = u^2 + 2as       where u is the initial velocity

0 = u^2 + 2*-5*15

u^2 = 150

so u = √150

Now we use v = u + at:

0 = √150 - 5t

t = √150/5

t = 2.45 seconds.

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2 years ago
Estimate 1 13 cos(x2) dx 0 using the Trapezoidal Rule and the Midpoint Rule, each with n = 4. (Round your answers to six decimal
babunello [35]

Answer:

(a) 4.152698

(b) 3.215557

Step-by-step explanation:

(a)

\int\limits^{13}_1 {cos(x^2)} \, dx =M_n=$\sum_{n=1}^{\infty} f(m_i)\Delta x $

n=4, so :

Each subinterval has length :

\Delta x= \frac{b-a}{n} =\frac{13-1}{4} =\frac{12}{4} =3

Therefore the subintervals consist of:

[1,5], [5,9], [9,13]

Now, the midpoints of these subintervals are:

\frac{1+5}{2} =3\\\\\frac{5+9}{2} =7\\\\\frac{9+13}{2} =11

Hence:

M_4= 3*(cos(3^2))+3*(cos(7^2))+3*cos((11^2))\approx 4.152698

(b)

\int\limits^{13}_1 {cos(x^2)} \, dx =T_n=\frac{\Delta x}{2} (f(x_o)+2f(x_1)+2f(x_2)+...+2f(x_n_-_1)+f(x_n))

n=4, so :

Each subinterval has length :

\Delta x= \frac{b-a}{n} =\frac{13-1}{4} =\frac{12}{4} =3

Therefore the subintervals consist of:

[1,5], [5,9], [9,13]

The endpoints of the subintervals consist of:

5,9

Hence:

T_4= \frac{3}{2}(cos(1^2)+2*cos(5^2)+2*cos(9^2)+cos(13^2)) \approx 3.215557

8 0
3 years ago
Please help ASAP!! please no thank you have a great day!
Archy [21]

1 and 5 are yes and the rest are no

7 0
3 years ago
Read 2 more answers
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