Question

in oigh 76.9 60 graph Based on these information answer questions 5 and 6 time () S. Find the average acceleration of the object from -2stor-8 a) 5,00 mvs b)) 3.33 m/s e) 7.60 d) -0.33 m/s 2 6. How far the object moves from to-0sto-8 a) 90 m b) 110 m c) 120m

Answer 1

5.

The velocity at t = 2 s is 10 m/s.

The velocity at t = 8 s is 30 m/s.

So, the average acceleration is

$a_{avg}=\frac{\Delta v}{\Delta t}=\frac{30-10}{8-2}=\frac{20}{6}=3.33\ m/s^2$

6.

The distance traveled is the area under the curve.

So, from t = 0 s to t = 8 s the graph can be divided into 3 parts.

i) t = 0 to t = 4

$d_1=\frac{4\times 20}{2}=40\ m$

ii) t = 4 to t = 6

$d_2=2\times 20=40\ m$

iii) t = 6 to t = 8

$d_3=2\times 20+\frac{2\times10 }{2}=50\ m$

So, the total distance covered during this time is