Question

IMPORTANT QUESTION! All help would be appreciated.

Answer 1

The formula for final velocity of an object starting from zero and undergoing constant acceleration is \[v=\frac{ 1}{ 2 }at^2\]
What is the time it will take a car accelerating at \[4 m/s^2\] to reach \[50 m/s\]?

Please help!

Answer 2

@whpalmer4
I hate to bother you but I'm a bit confused...

Answer 3

@ganeshie8

Answer 4

Uh, I'm not so sure you have your formula correct. velocity has units of m/s, but that formula multiplies acceleration (m/s^2) by time squared (s^2) so you get m as the final unit.

Answer 5

So you are right to be confused :-)

Answer 6

I do A Plus though, it's a school program kind of "go at your own pace" type thing.

Answer 7

Now, perhaps you meant \[v_f = v_i + at\]which gives you the final velocity \(v_f\) given starting velocity \(v_i\) (0 in your case) and constant acceleration \(a\) for time \(t\)

The formula you gave is very close to the formula for position of an object from a standing start with constant acceleration: \[x = \frac{1}{2}at^2\]

Answer 8

The question says the formula I wrote in the question :/ It may be wrong, but even though the equation may be wrong, this is what I plugged in for it.
\[50=\frac{ 1 }{ 2 }(4)(t)\]

Answer 9

The question is incorrect. You can't just use the numbers, the units must also make sense! They are just as much an integral part of the problem as the numbers.

Here, I'll show you how to do the problem correctly:

\[v_f = v_i + at\]\[v_i = 0\]because we are starting at rest.
\[v_f = at\]\[v_f = 50\text{ m/s}\]\[a = 4\text{ m/s}^2\]
\[\frac{v_f}{a} = \frac{at}{t}\]\[t = \frac{v_f}{a} = \frac{50 \text{ m/s}}{4 \text{ m/s}^2} = \frac{50}{4}*\frac{\frac{m}{s}}{\frac{m}{s^2}} = 12.5 \text{ s} \]

Answer 10

Once again, thank you so much for your help and patience. It's extremely appreciated!

Answer 11

Here's a table showing time, velocity, and distance traveled by a car starting at rest and accelerating 4 m/s^2, as in your problem:

\[\begin{array}{ccc}
t & v & x\\\hline\\
0 & 0 & 0 \\
1 & 4 & 2 \\
2 & 8 & 8 \\
3 & 12 & 18 \\
4 & 16 & 32 \\
5 & 20 & 50 \\
6 & 24 & 72 \\
7 & 28 & 98 \\
8 & 32 & 128 \\
9 & 36 & 162 \\
10 & 40 & 200 \\
11 & 44 & 242 \\
12 & 48 & 288 \\
13 & 52 & 338 \\
\end{array}\]

Answer 12

A bit of calculus without tears:

Answer 13

Notice in the graph that the purple line (acceleration) is constant. The velocity line (olive) is slanted, but straight. The position line (green) is a parabola — in fact, it is direct variation of position with the square of time, and constant of variation is \(\frac{1}{2}a\). I remember we did such a problem yesterday :-)

Answer 14

Now, calculus is all about rates and adding things up. The position of a moving object is just its prior position, plus the amount it moved, right? If your car is at x = 0 at t = 0, and it isn't moving, at t = 1, it will still be at x = 0, right? If your car is moving at v = 1 unit/s, then at t = 0 it is at x = 0, but at t = 1, it is at x = 0 + v*t = 0 + 1 unit/s*1 s = 1. At t = 2, it will be at x = 1+1 = 2. And so on. That's constant velocity (speed, really) motion.

Now, if it is moving at a constant speed, then if we graph the position of the car, it is a straight line, with the slope of the line corresponding to the speed. Your car moving at a speed of 1 m/s will have an equation of x = vt or x = 1t. There is no acceleration, so the speed remains the same.