Physics...acceleration and velocity problem?

Question

a drag racer, starting from rest, speeds up for 402m with an acceleration of +17.0 m/s^2. A parachute then opens, slowing the car down with an acceleration of -6.1m/s^2. how fast is the racer moving 350m after the parachute opens?

Answer 1

The working formula is

Vf^2 - Vo^2 = 2as

where

Vf =  final velocity

Vo = initial velocity

a = acceleration

s = distance travelled

First step is to determine the velocity of the car just before the parachute opens,

Vf^2 - Vo^2 = 2(17)(402)

Since the car started from rest, then Vo = 0, hence

Vf^2 = 13668

Vf = 116.91 m/sec.

To determine the speed of the car when it has moved 350 meters after the parachute opens, use the same formula used above

Vf^2 - Vo^2 = 2as

but this time, the variables are defined as follows:

Vf = velocity of the car after moving 350 m

Vo = velocity of the car before the parachute opens = 116.91 m/sec (as calculated above)

a = -6.1 m/sec^2 (given)

s = 350 m (given)

Substituting appropriate values,

Vf^2 - 116.91^2 = 2(-6.1)(350)

Vf^2 = 116.91^2 - 2(6.1)(350)

Vf^2 = 9398

Vf = 96.94 m/sec.

Hope this helps.

Answer 2

sorry i forgot

Answer 3

There are 2 ways to do this.

1. Use the basic equation of motion for distance and solve for t:

d = d0 + v0t + ½at²  

d0 and v0 are both 0, therefore:

t = √2d/a

Then, use the equation of motion for speed

v = at

substitute t from above:

v = a √2d/a

= √2da²/a

= √2da

= √2 (402) (17)

= 116.91 m/s after 402 m

2. You can also compute the speed after 402 m by this equation:

v² = u² + 2ax

u is 0, thus

v² = √2ax

v = √2*17.1*402  

v = 116.910 m/s after 402 m

The same equation works for deceleration:

v² = u² + 2ax

u = speed when chute applied

v = speed after 350 m

v² = u² +2 (-6.1) (350)

v = √116.910² - 4270

= √13667.9481 - 4270

= 96.94 m/s after 350 m

I think I did the math right. Be sure to double check.