Answer:
acceleration of the car is 3 m\s^2
Explanation:
from rest means the initial velocity (vi) is zero
time = 5s
final velocity (vf) = 15m\s
a = vf - vi \ t
a = (15-0) \ 5
a= 3 m\s^2
which means that the car is speeding up 3 meters every second
The time that the car take to catch up to truck would be :
1/2 ar^2 = vt
1/2 (3) t^2 = 15t
1.5 t^2 = 15t
t ^2 = 10s
Hope this helps
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Answer:
v=1.295
Explanation:
What we are given:
a=5÷(3s^(1/3)+s^(5/2)) m/s^2
Start by using equation a ds = v dv
This problem requires a numeric method of solving. Therefore, you can integrate v ds normally, but you must use a different method for a ds The problem should look like this:

<em>a=2</em>
<em>b=1</em>
<em>x=5÷(3s^(1/3)+s^(5/2)) </em><em>m/s^2</em>
<em>dx=dv</em>
Integrate the left side the standard method.

<em>a=v</em>
<em>b=0</em>
<em>dx=dv</em>
<em>Integrating</em>
=v^2/2
Use Simpson's rule for the right site.

<em>a=b</em>
<em>b=a</em>
<em>x=f(x)</em>
f(x)=b-a/6*(f(a)+4f(a+b/2)+f(b)
If properly applied. you should now have the following equation:
v^2/2=5[(1/6*(0.25+4(0.162)+(0.106)]
=0.8376
Solve for v.
v=1.295