Question

The acceleration of a cart started at t =0, varies with time as shown in figure (3-E2). Find the distance travelled in 30 seconds draw the position-tome graph.

Answer 1

In fiirst 10 sec,$S_1=ut+\frac{1}{2}a{t}^2$
=0+$\frac{1}{2}{5.10}^2=250ft$
v=u=at
=0+$5\times 10=50\frac{\text{ft}}{\text{sec}}$

$\therefore$ From 10 to 20 sec $\Delta$=20-10=10 sec) it moves with uniform velocity of 50 $\frac{\text{ft}}{\text{sec}})$
Distance, $S_2$=$50\times 10=500ft$
Between 20 sec to 30 acceleration is constant i,e.-5$\frac{\text{ft}}{\text{s^2}}$

At 20 sec, velocity is 50$\frac{\text{ft}}{\text{sec}}$
t=30-20=10 s

$S^3$=ut+$\frac{1}{2}a{t}^2$
50$\times 10+\frac{1}{2}(-5){10}^2$
=500-250=250 m

Total distance travelled is 30 sec

=$S_1+S_2+S_3$
=250+500+250
=1000 ft