Velocity/Acceleration problem?

Question

A car starts from rest and travels for 5.0 s with a uniform acceleration of +2.7 m/s2. The driver then applies the brakes, causing a uniform acceleration of -3.0 m/s2. If the brakes are applied for 2.0 s, determine each of the following.

(a) How fast is the car going at the end of the braking period?

(b) How far has it gone?

Answer 1

You know sometimes it's easier to visualize the situation graphically, rather than using abstract formulae.

1. Plot speed versus time on the acceleration phase, and you see that after 5 seconds speed has risen from 0 to 13.5 m/s just prior to braking. The area of the right triangle describing speed is 0.5(b)(h) or 0.5(5)(13.5) or 33.75 meters, the distance travelled during acceleration. You have just integrated speed to arrive at distance.

2. Next continue the plot of velocity with a downward slope of 3m/s^2 for 2 seconds. At the end of two seconds, speed has dropped from 13.5 m/s to 7.5 m/s which is the speed at the end of the braking period.

3. Again ntegrate your speed/time plot during the braking event to get distance. The area formed by the 2 second deceleration by 6 m/s is 6m, while the area formed by the residual velocity of 7.5m/s over 2 seconds is 15m.

Summing 33.75m+6m+15m you get that the car has travelled 54.75m.

Answer 2

Speed reached and distance covered for the 5 sec acceleration. Starts at speed = 0 (at rest)

Speed = V2 = acceleration x time ( 2.7 m/sec^2) x 5 sec

Distance traveled during acceleration = xa = 1/2 acceleration x (time^2)

During braking the care decelerates from V2 to V? and the deceleration rate is - 3.0 m/s^2 for a time = 2 sec

V? = V2 - (3.0) x (2) final velocity

braking distance = xb = V2 x 2 sec - 1/2 (3.0)*(2)^2

Total distance = xa + xb

thats it.

Answer 3

If the acceleration is constant .the velocity = acceleration*time+ initial velocity
In this case,after 5 s the velocity is 5*2.7 =13.5m/s as the car starts from rest
The same formula applies during braking
v=a*t +v_i in this case a=-3m/s^2 and v_i= 13.5 m/s
v=-3*2+13.5 = 7.5 m/s

The formula for the space gone is

space = 1/2 a*t +v_i*t + space at beginning

we divide the whole period into two

1)Acceleration period space = 1.35*5^2 = 33.75 m
2)Braking period

space = -1.5*4 +13.5*2 + 33.75 = 54.75 m which is how far it has gone

Answer 4

Firstly determine the speed to which the car climbs before decelerating, v=u+at=>v=(2.7)(5)m/s=13.5m/s (initial velocity of 0m/s)
Calculate v, given that u is now 13.5m/s (the new initial velcocity), the acceleration is -3.0m/s^2, t=2s => v=u+at=13.5+(-3.0)(2)=7.5m/s
Hence the answer to (a) is 7.5m/s.
The distance travelled in the first instance (before breaking) is s=0.5at^2=0.5*2.7*5^2=33.75m. While decelerating the distance travelled is u^2/2a (since v is 0) =>s=13.5^2/2(3.0)=30.375m
Therefore, the overall distance travelled is 33.75m+30.375m=64.125m, the answer for (b).