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The cart started at t=0,its acceleration varies with time as shown in figure. Find the distance travelled in 30 seconds and draw the position time graph.

The cart started at t=0,its acceleration varies with time as shown in figure. Find the distance travelled in 30 seconds and draw the position time graph.

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2 Answers

Susmita
425 Points
5 years ago
In t=0 to 10 s,acceleration f=10 m/s2.Initial velocity u=0 at t=0.Final velocity v at t=10s can be calculated from
v=u+ft
Or,v=10×10=100 m/s
Distance s travelled in this time can be calculated from
v2=u2+2fs
or,s1=v2/2f=1000/2=500m
In the time interval t=10 to 20s,f=0.So particle moves with constant speed v=100m/s.So distance travelled is s2= vt=100×10=1000m
In the time interval t=20 to 30 second,f=-10m/s2.Initial velocity is u=100m/s and final velocity
v=u+ft
Or,v=100-100=0m/s
Distance travelled
v2=u2+2fs3
Or,s3=500 m.
So total distance travelled is 2000.
To draw the graph for the 1st segment,
f=10
Or,d2x/dt2=10
Integrating
dx/dt=10t+c
Velocity is zero at t=0.so c=0.
Again integrating
x=10t2/2=5t2
It is a parabola
For the second segment,
f=0
d2x/dt2=0
Or,dx/dt=constant=say c
Integrating
x=ct+c1
This is a straight line.
For the 3rd segment
x=-5t2+c2t+c3
It is also a parabola.
Please try to draw these in sequentially in a graph as I have no option of attaching file from my device.
Hope this helps.
Susmita
425 Points
5 years ago
In both cases draw half the parabola.Beacuse t is only positive here.
In the 1st case the parabola is of the form
y=x2.Draw only the right side of it starting from  origin.
In the 3rd segment it is of the form y=-x2.In this case also draw only the right half of the parabola.

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