Velocity and Acceleration

**Aminekhadir** · December 3rd 2009, 03:50 AM

Hello!
This question is about the function:
v = 50 - 50e^(-0.2t), t is bigger or the same as 0.
v is in metres per second and t is in seconds.

The problem is to find:
a) i- an expression for the acceleration, a, as a function of t.
ii - What is the value of a when t = 0?

also,
d) Let y metres be the distance fallen after t seconds.
i) show that y = 20t + 250e^(-0.2t) + k, where k is a constant.

I would appreciate your help,
best wishes Amine

**dedust** · December 3rd 2009, 05:35 AM

(i) $v=50 - 50e^{-0.2t}$ and $a=\frac{dv}{dt}$

(ii) when you get $a$ as a function of $t$ , put $t = 0$

(iii) $v=\frac{dy}{dt}$ , so $y=\int v dt=\int50 - 50e^{-0.2t} dt$

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