Question

An object travels as the description shown below.

From t = 0 s to t = 7 s, the object was travelling with a constant velocity of 20 m/s. From t = 7 s to t = 11 s, it underwent a constant acceleration and reached a final velocity of 40 m/s at t = 11 s. The object is travelling to the right for the entire motion.
a) What was the acceleration of the object at t = 8 s?
b) What was the value of t when the object’s speed was 28 m/s?

Answer 1

@JoannePo are you here?

Answer 2

a) since velocity is not constant at second period, we use the following formula:

\[v_f=v_i+0.5at^2\]
\[a=2(v_f-v_i)/t^2\]
\[a=2(40-20)/(4)^2\]
\[a=40/16=2.5 \space ms^{-2}\]

b) use the same formula to find t at vf=28 m/s (your homework)

Answer 3

i'm here

Answer 4

a). time taken(11-7) and velocity changed (40-20)
so acceleration which is contant would be velocity over time
thats is 20/4=5ms^-2 at any instant the aceleration would be 5 as its constant
b)
t as 20=7
after scceleration at t=8 speed would be 25
after t=9 speed would be 30
so the time taken would be difference in velocity/acceleration that is 8+(28-25)/5
that is 8+3/5 or 8+3/5*60sec or 8min 36seconds