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 and draw the position-time graph.

Answer 1

Answer:

whaere is the figure send the pic of the figure also

Answer 2

Distance travelled in 30 seconds, 1000 m .

Explanation:

Given data in the question :

          a = 5 ft/s²          

 Time (t) = 10 second

        $s_{1}=u t+\frac{1}{2} a t^{2}$

        $s_{1}=0 \times 10+\frac{1}{2} \times 5 \times 10^{2}$  

        $s_{1}=0+\frac{1}{2} \times 5 \times 100$

        $s_{1}=5 \times 50$          

        $s_{1}=250 \mathrm{ft}$                          

For next 10 s, a = 0

        $s_{2}=v t$

        $s_{2}=50 \times 10$

        $s_{2}=500 \mathrm{ft}$

For next 10 s, a = -5 ft/s²

        $s_{3}=u t+\frac{1}{2} a t^{2}$

        $s_{3}=50 \times 10+\frac{1}{2} \times(-5) \times 10^{2}$

        $s_{3}=500+\frac{1}{2} \times(-5) \times 100$

        $s_{3}=500+(-5) \times 50$

        $s_{3}=500-250$

        $s_{3}=250 \mathrm{ft}$

Average distance covered in 30 s,

         $s=s_{1}+s_{2}+s_{3}$

         s = 250 + 500 + 250

         s = 1000 ft

         s = 300 m

Thus, the total distance traveled is 1000 ft.