physics problem help? please?

Question

ok, so i tried everything and decided to come here

1. a car starts from rest and travels for 5.0s with a uniform acceleration of +1.9m/s^2. the driver then applies the brakes, causing a uniform acceleration of - 1.6 m/s^2.
A) if the brakes are applied for 2.6 s, how fast is the car going at the end of the braking period? answer: 5.34 m/s
B) how far has it gone from it's start? answer in units of m
it is not 9.5m or 13.66m

?? please help

and

a plane lands with a velocity of +110.3m/s and accelerates at a maximum rate of -5.5m/s^2
A) from the instant the plane touches the runway, what is the minimum time needed before it can come to rest? answer: 20.054
B) the plane is landing on a naval aircraft carrier that is 0.80 km long. what distance does the plane require to land? answer in units of km

it is not 0.1103km or 0.60665km

yes it is help on my homework that is why it is homework help
please any help would be appreciated

well i really can't do that since i am in high school and have to submit the answers by 6am and can't ask my teacher because it is the middle of the night

Answer 1

Question 1.

use this formula

v = u +ft (v - final velocity, u - initial velocity, f - acceleration, t - time)

now while accelerating for 5 seconds the car has attended a certain speed let's say v1 and the v1 = 0 + 1.9X5 meter/second
or, v1 = 9.5 meter/second

now deceleration is applied for 2.6 seconds so let's say the speed becomes v2

then v2 = v1 - ft (negative sign as it is deceleration)
v2= 9.5 - 1.6X2.6 = 9.5-4.16=5.34 meter/second

(A) So the car will be moving at 5.34 meter/second after applying brake

To see the distance use this formula

S = ut + 1/2 ft^2 (s - distance travelled, u - initial speed, f - acceleration, t - time)

Lets say the car moves s1 meters while accleration and moved s2 meters while deceleration and addition of s1+s2 will give us the total distance travelled.

so, s1 = 0X5 + 1/2 (1.9)(5)^2 = 1/2X1.9X25 = 23.75 meters

and s2 = 5.34X2.6 - 1/2 (1.6)(2.6)^2 ( negative sign as it is deceleration)

so s2 = 13.884 - 1/2 X1.6X6.76 = 13.884-5.408 = 8.476 meters

so total distance travelled is 23.75+8.476 = 32.226 meters

Problem 2:

Same formula

v = u - ft (negative as it is deceleration)

0 = 110.3 - 5.5 t
or, t = 110.3/5.5 = 20.0545 seconds (The plane will come to stop within 20.0545 seconds) --- (A)

okay now use this formula

s = ut - 1/2 f t^2 (negative as it is deceleration)
s = 110.3X20.0545 - 1/2 (5.5)(20.0545)^2
s = 2212.016 - 1106.008
s = 1106.08 meters
s = 1.10608 km

If this plane lands on naval aircraft of 0.8 km length, it can not do so, as the applied deceleration will pull it beyond the landing stip.

Hope this helps

anymore help send me an email I will try to help you out :) cheers

Answer 2

Consult a Physics who has got a degree in Physics and like to teach by heart as well.