# Acceleration Problems Calculus

[mage lang="" source="flickr"]acceleration problems calculus[/mage]

Find the equation of the line tangent to
y = e^(x+1) at x= 0

The second problem is this:

acceleration: a(t)= 2e^t at t = 1
velocity: v = 2

find v(0)

Thanks!!

First of all, we need to know the equation of a line. The equation of a line is y – y1 = m(x – x1) where (x1, y1) is a point that falls on the line and m is the slope of the line. We know that x = 0, so we can replace x1 with 0:

y – y1 = m(x – 0)

We need to find y1 and m. We can find y1 easily by plugging into the equation y = e^(x+1) when x = 0:

y(0) = e^((0)+1)
y(0) = e

Substituting y1 with e, we get:

y – e = m(x – 0)

We still need to find m. The slope of the line is the derivative of this equation evaluated at x = 0:

dy/dx = e^(x+1)
dy/dx @ x=0 = e^((0)+1) = e

Now we substitute m with e and the equation of the line tangent is:

y – e = ex

For the second problem, acceleration and velocity are related in that the derivative of velocity is acceleration and the integral of acceleration is velocity. I’m not sure how you got v = 2. Also we can’t solve the problem unless we have an initial condition, like v(a) = c.

Calculus – Position, Velocity and Acceleration

Category: Calculus Problems
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