a) What is the maximum acceleration the car can achieve? I used the equation a =

Question

a) What is the maximum acceleration the car can achieve?

I used the equation a = us(g). Is this correct? I got 7.84 m/s^2.



b) And if the initial speed down is 26.7 m/sec, how far (distance) does the car require to stop?

For this I used xf = xo + vo(x direction) * sin20 and got 9.13 m/s.

c) And what is the car's acceleration and how far (distance) does it take for the car to stop if it is now driving the same slope uphill.

I'm not understanding if this will be different from our answer in b, would the direction just have a - in front of our velocity?

Explanation / Answer

unfortunately your answers are wrong!
a) let me to start with the forces:

in this case the friction force is against the movement direction and the weight component force is in the same direction with movement. So:
ma= mgsin(20) - (mg)us
we can cancel m from equation
we have:
a=g sin(20)- g us= 3.35-7.84= -4.488 m/s^2

b) you should use this formula
(V2)2-(V1)2=2ax

0- 26.72= 2*-4.488*X

X= 79.42m

c)

we have:

ma= -mg sin(20) - (mg)us

a= -g sin(20) - (g)us

a= -11.19

(V2)2-(V1)2=2ax

0- 26.72= 2*- -11.19*X

X= 31.85

 

 

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