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

Answer

A

Explanation

In first `10 sec
S_1 = ut+1/2 at^2
=0+1/2 5.10^2=250ft.
At 10 sec
v=u=ast
=0+5xx10=50 ft/sec. :.` From 10 to 20 sc `(/_\t=20-10=10)` it moves with uniform velocity of 50 ft/sec.
Distance
`S_2=50xx10=500 ft`
Between 20 sec to 30 sec acceleration is constant i.e. `-5 ft/s^2

At 20 sec
velocity is 50 ft/ sec
`t=30-20=10 s
S_3= ut+ 1/2 at^2
=50xx10+1/2(-5)10^2
=500-250=250m`
Total distance travelled is 30 sec `=S_1+S_2+S_3
=250+500+250
=1000 ft.`