can you help with acceleration?

Question

At the end of a run on a quarter-mile track the parachute opens and the car slows from 210 South to 10 mph in 10 secs. What was the average acceleration of the car during that 10-sec parachute breaking period?

Answer 1

Acceleration is a change in velocity with respect to time.
Acceleration = Δv / Δt

The change in velocity can be calculated as the final velocity of the car minus the initial velocity,
Δv = v_final – v_initial

The initial velocity of the car was 210 mph South.
The final velocity of the car was 10 mph South.
The change in the velocity was thus, -200 mph South, or +200 mph North.

The time it took for the velocity to change, we are told, is 10 seconds.
Δt = 10 second.

So we can find the acceleration to be,
a = Δv / Δt
a = -200 mph South / 10 seconds
a = -20 mph per second
Which is the same as +20 mph per second North

The direction is important since acceleration (like velocity) is a vector and has not only a magnitude but also a direction.
Since the North and South directions act in parallel / anti-parallel directions (i.e. there is no perpendicular component of one with respect to the other), we can represent North as the opposite direction as South. Any vector pointing South can be represented as a vector of equal magnitude, but opposite sign, pointing North. For example, -200 mph South = +200 mph North.

Answer 2

The velocity changes from 210 mph to 10 mph in 10 seconds.

The formula for acceleration is

v = u + at where u is the initial velocity, v is the final velocity, a is the acceleration (or deceleration) and t is the time.

So a = v - u / t

Let us use the mph values and convert them into feet per second at the same time.

a = [(10 - 210) X 1760 X 3] / [ 3600 X 10]

= (- 200 X 1760 X 3) / (3600 X 10)
= -1760 / 60 = -176 / 6 = - 29.333 ft/sec.sec

Since we have a slowing down, a is -ve.

Answer 3

there is alot extra in the back of who will win .. a/c will will enhance up linear.. or otherwords, a continuing fee of acceleration(with some small fluctions using pull of the wheels, flaps ) till the thrust=drag. autos have Gears, using fact of this acceleration isn't consistent ordinary. each kit has a diverse acceleration fee, till the max potential obtainable is acheived. 0.33 is potential to weight ratio- the extra merchandise weighs, the extra rigidity is had to enhance up it. Physics one hundred and one, F=MA In a quick distance, a automobile might win, only using time required for the engines to spool up, and enhance up the airplane, yet over an prolonged distance, the acceleration from the airplane will over take the automobile. The quickest highway criminal automobile is recorded at 253mph (Buggati Veyron) . A b747 flies at around 380kts

Answer 4

So in plain English;
The difference from 210Mph to 10Mph is 200Mph
200 / 10 = 20
So for every second the car slows down 20 Mph

Answer 5

acceleration = change in velocity / change in time
= (10 mph - 210 mph)/ 10 sec x (3600 sec / hr) = -72,000 miles / hr^2

Answer 6

The calculations given by everyone else (assuming they are right, I didn't check their work) are AVERAGE acceleration. The actual acceleration is rapidly changing in as the velocity changes! check out the sources for more info on the beautiful and complicated world of (fluid) mechanics!

Answer 7

if you want deceleration in m/hr^2 change the seconds to hours

10 s= 10/3600=2.78E-3
The deceleration is (10-210)/2.78E-3=-72E3 mi/hr^2 South
or +72E3 mi/hr^2 North

Answer 8

acceleration = (final velocity - initial velocity)/time

convert 10mph and 210mph to meters per second

210 mi/h * 1.6 km/mi * 1000 m/km * 1 h/60min * 1 min/60s
= 93.33 m/s

10 * 1.6 km/mi * 1000 m/km * 1 h/60min * 1 min/60s
= 4.44 m/s

(4.44 m/s - 93.33 m/s) / 10s
= -8.89 m/s^2 or the parachute decelerated by 8.89 m/s^2

Answer 9

Accerelation - velocity / time.
200/10 = - 20miles/ s

Answer 10

i could be incorrect, but i think there is no acceleration during that 10 second period of deceleration. but i am no mathematician