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hodyreva [135]
3 years ago
13

a car traveling at 28 m/s slows down at a constant rate for 4 seconds until it stops. what is its acceleration?

Physics
1 answer:
astra-53 [7]3 years ago
3 0
We are given the following conditions:
v_{0} = 28m/s

t = 4s

v_{f} = 0m/s

Since we are told it slows down at a constant rate, we know we are dealing with uniformed accelerated motion, or UAM for short.

So we look at our kinematics equation that contains the given variables.

v_f = v_0 + at

This contains our target variable and our initial conditions. Plug and chug, we get:

(0m/s) = (28m/s) + a(4s) -28m/s = (4s)*a a = -7m/s^2

Thus the answer is:

a = -7m/s^2

Hope this helps!


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The terminal velocity is mathematically represented as

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      v_t  =  \sqrt{\frac{2 * 0.510 *  9.8 }{1.21 * 1  * 0.026912 } }

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