How do I calculate distance from acceleration and velocity?

[Eclair_de_XII]

Homework Statement

"An electric vehicle starts from rest and accelerates at a rate of ##2.0 \frac{m}{s^2}## in a straight line until it reaches a speed of ##20 \frac{m}{s}##. The vehicle then slows at a constant rate of ##1.0 \frac{m}{s^2}## until it stops. (a) How much time elapses from start to stop? (b) How far does the vehicle travel from start to stop?"

Homework Equations

##v = v_0+at##
##x-x_0=v_0t+\frac{1}{2}at^2##
Answer to (b): 300 m

The Attempt at a Solution

(a)
##20\frac{m}{s} = 2\frac{m}{s^2}(t_1)##
##t_1=10s##
##0\frac{m}{s}=20\frac{m}{s}-1\frac{m}{s^2}(t_2)##
##-1\frac{m}{s^2}t_2=-20\frac{m}{s}##
##t_2=20s##
##t=t_1+t_2=10s+20s##

(b)
##x_1-0m=(0\frac{m}{s})(10s)+\frac{1}{2}(2\frac{m}{s^2})(10s)^2##
##x_1=100m##
##x_2-100m=(20\frac{m}{s})(20s)+\frac{1}{2}(-1\frac{m}{s^2})(20s)^2##
##x_2=100m+400m-200m=300m##
##x=x_1+x_2=100m+300m ≠ 300m##

 

[ehild]

Homework Statement

"An electric vehicle starts from rest and accelerates at a rate of ##2.0 \frac{m}{s^2}## in a straight line until it reaches a speed of ##20 \frac{m}{s}##. The vehicle then slows at a constant rate of ##1.0 \frac{m}{s^2}## until it stops. (a) How much time elapses from start to stop? (b) How far does the vehicle travel from start to stop?"

Homework Equations

##v = v_0+at##
##x-x_0=v_0t+\frac{1}{2}at^2##
Answer to (b): 300 m

The Attempt at a Solution

(a)
##20\frac{m}{s} = 2\frac{m}{s^2}(t_1)##
##t_1=10s##
##0\frac{m}{s}=20\frac{m}{s}-1\frac{m}{s^2}(t_2)##
##-1\frac{m}{s^2}t_2=-20\frac{m}{s}##
##t_2=20s##
##t=t_1+t_2=10s+20s##

(b)
##x_1-0m=(0\frac{m}{s})(10s)+\frac{1}{2}(2\frac{m}{s^2})(10s)^2##
##x_1=100m##
##x_2-100m=(20\frac{m}{s})(20s)+\frac{1}{2}(-1\frac{m}{s^2})(20s)^2##
##x_2=100m+400m-200m=300m##
##x=x_1+x_2=100m+300m ≠ 300m##

x_2 is the position, not the displacement during the second stage of motion.

 

[akshay86]

you have mistaken in attempting question b
only x₂-100 gives the displacement in the time of disceleration and x₂ is the total displacement and you calculated it has 300m(correct)

 

[PeroK]

Homework Statement

"An electric vehicle starts from rest and accelerates at a rate of ##2.0 \frac{m}{s^2}## in a straight line until it reaches a speed of ##20 \frac{m}{s}##. The vehicle then slows at a constant rate of ##1.0 \frac{m}{s^2}## until it stops. (a) How much time elapses from start to stop? (b) How far does the vehicle travel from start to stop?"

With a problem like this (two different acclerations) it can be useful to draw a graph of the motion: velocity against time. The displacement is then the area under the graph. It helps you visualise the motion and can prevent you from plugging the wrong numbers into your equations.

 

[Eclair_de_XII]

only x2-100 gives the displacement in the time of disceleration and x2 is the total displacement and you calculated it has 300m(correct)

So you're saying that x2 = x1 + x3, where x3 is some arbitrary distance between the end of x1 and x2? Then that would mean...

##x_2-100m=(20\frac{m}{s})(20s)+\frac{1}{2}(-1\frac{m}{s^2})(20s)^2 = x_1+x_3-x_1 =x_3##
##x_3=(20\frac{m}{s})(20s)+\frac{1}{2}(-1\frac{m}{s^2})(20s)^2=400m+(-200m)=200m##
##x_2=x_3+x_1=200m+100m=300m##

Thanks, guys.