Anonymous
Anonymous asked in Science & MathematicsPhysics · 10 years ago

# Physics Help! 10 Points!!!!?

Ok, I am stuck on these! Any help is appreciated, I will pick a best answer asap.

1) An electric vehicle starts from rest and accelerates at a rate of 2.0 m/s2 in a straight line until it reaches a speed of 16 m/s. The vehicle then slows at a constant rate of 1.2 m/s2 until it stops. (a) How much time elapses from start to stop? (b) How far does the vehicle move from start to stop?

I got part a correct, which is 21.33 seconds, but am confused on b.

2) A certain elevator cab has a total run of 191 m and a maximum speed is 312 m/min, and it accelerates from rest and then back to rest at 1.15 m/s2. (a) How far does the cab move while accelerating to full speed from rest? (b) How long does it take to make the nonstop 191 m run, starting and ending at rest?

3) A car traveling 59.2 km/h is 21.9 m from a barrier when the driver slams on the brakes. The car hits the barrier 2.31 s later. (a) What is the car's constant deceleration magnitude before impact? (b) How fast is the car traveling at impact?

Relevance
• Anonymous
10 years ago

the first :

the car is accelerating on a straight road , so we will use the formulla : x=x0 +v0*t +0.5*a*t^2 .

to calculate the time passed use: v=v0+at >>> 16=0+2t >>>> t=8sec

0 = 16 -1.2t >>>> t=40/3 >>>> total time = 8 +40/3 .

distance 1 : x = 0 + 0 +0.5 * 2 *8^2 = 64m

distance 2 : x= 0 + 16*(8+40/3) - 0.5*1.2*(8+40/3)^2 = please calculate it .

total distance = dis1+dis2

sorry , i cant understand the second because i am not that good in english

the third :

first we will convert the speed units to m/sec : 59.2 km/h = 59.2/3.6 m/sec = (16+4/9) m/sec

then use x=x0+v0*t +0.5*a*t^2 >>>

0=-21.9 + (16+4/9)*2.31+0.5*a*(2.31)^2 >>>> a=-6.02937 m/sec^2

using the momentum constant , the speed after impact is equivalent to the speed the time the car hits the barrier ....

so : we will find the speed of the car when is hit the barrier :

v=v0+a*t >>> v = (16+4/9) + -6.02937*2.31 =2.51660

and this is the impact speed but with different sign , because the car moves on the different direction , means -v = -2.51660

hope i was useful to you mate

• Part b) is easy if you use the equation " X = Xo + Vot + (1/2)at^2".

For the first part when the vehicle is accelerating the total time was 8 sec, which you solved.

We now have the known variables of:

Xo = 0

Vo = 0

a = 2 m/s^2

t = 8 sec

Using the equation input the known variables and solve for X:

X = Xo + Vot + (1/2)at^2

X = 0 + 0 + (1/2)(2 m/s^2)(8^2)

X = 64 m

For the second part where the vehicle is decelerating the total time was 13.3 sec, which you also solved.

The known variables are:

Vo = 16 m/s

Xo = 64 m

a = -1.2 m/s^2

t = 13.3 sec

Using the equation input the known variables and solve for X:

X = Xo + Vot + (1/2)at^2

X = 64m + 16 m/s(13.3s) + (1/2)(-1.2 m/s^2)(13.3s)^2

X = 170.6 m

So the total distance is near 171 meters.