A car slows down uniformly from a speed of 22.0 m/s to rest in 5.50 s. how far did it travel in that time?

2 answers:

8 0

Since the acceleration is 'uniform', the car's average speed during that time is 11 m/s.

Traveling at an average speed of 11 m/s for 5.5 sec, it covers

(11 m/s) x (5.5 sec) = 60.5 meters. (No calculus. Hardly any algebra. Mostly arithmetic.)

svetlana [45]3 years ago

3 0

First get the acceleration. Since it's uniform we use

$a= \frac{-22m/s}{5.5s}=-4 \frac{m}{s^2}$

Note the acceleration is negative since we start at a positive speed and end at zero.
Now the distance is the acceleration integrated twice. The first integral gives the velocity at any time, t

$v=- \int\limits^{ } _ {}{4 \frac{m}{s^2} } \, dt =-4t \frac{m}{s^2}+22.0\frac{m}{s}$

Notice if you put 5.5s for t in the expression we get 0 m/s as we should. Now to get the distance it traveled over this time we integrate this velocity expression:

$s= \int\limits^{5.5} _0{-4t \frac{m}{s^2} +22.0\frac{m}{s} } \, dt =-2t^2\frac{m}{s^2} +22.0t\frac{m}{s} \\ \\ Evaluating: \\ \\ -2(5.5)^2+22(5.5)=60.5m}$

So it travels 60.5 meters

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a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?

julsineya [31]

Answer:

Approximately $7.0\; \rm m \cdot s^{-1}$ .

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant $g = 9.81 \; \rm m \cdot s^{-2}$ near the surface of the earth.

For an object that is accelerating constantly,

$v^2 - u^2 = 2\, a \, x$ ,

where

$u$ is the initial velocity of the object,
$v$ is the final velocity of the object.
$a$ is its acceleration, and
$x$ is its displacement.

In this case, $x$ is the same as the change in the ball's height: $x = 2.5\; \rm m$ . By assumption, this ball was dropped with no initial velocity. As a result, $u = 0$ . Since the ball is accelerating due to gravity, $a = 9.81\; \rm m \cdot s^{-2}$ .