Question

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Answer 1

A car travelling at 72km/h decelerates uniformly at 2m/s^2.
Calculate
(a)the distance it goes before it stops
(b)the time it takes to stop
(c)the distance it travels during the first and third seconds

Answer 2

@LarsEighner

Answer 3

uniformy means its linear

Answer 4

so 2m/s is the slope
x(72)=0

so u have a point and slope, write the equation of distance x and speed v

Answer 5

No , i want to solve it using equations of motion

Answer 6

oh well not expert :P

Answer 7

Okay, if you are even vaguely familiar with calculus:

acceleration is the second derivative with respect to time
velocity is the first derivative with respect to time.
displacement is basic function.

you have

acceleration is -2 m/s^2

integrate that and you have

velocity(t) = 72km/h + (-2 m/s^2)t

Please do the conversion here. The metric system makes my skin crawl.

and

displacement(t) - (72 km/h)(t) + (1/2)(-2 m/s)(t) + initial displacement.

If you ever get confused, use dimensional analysis to check.

displacement is length
velocity is length/time
acceleration is length/time^2

Answer 8

You r great , i will be reading this , i am familiar with that not to worry

Answer 9

Linear (classical) motion cheat sheet:

\[ \large \int adt = at + C \]

The antiderivative of acceleration is velocity with respect to time. The constant \(C = v_0\), the initial velocity.

The dimension of acceleration is length / (time x time). The integral of acceleration with respect to time (i.e velocity) therefore has dimensions of length/ time.

\[ \large \int (at + v_0)dt = {1 \over 2}at^2 + v_0t + C \]

The antiderivative of velocity with respect to time is displacement. s is often used for displacement. Do not mistake this s for "speed." Use d in your own work to avoid confusion. In this case \(C = s_0 \) the original displacement

The dimension of displacement is length.

Going the other way:

\[ \large {d \over {dt}} s(t) = v(t) \]
\[ \large {d \over {dt}} v(t) = a \]