Question

The velocity vs. time graph for an object moving along a straight path is shown in the figure below.(m/s)Find the average acceleration of this object during the following time intervals.(a) 0 s to 5.0 sm/s²(b) 5.0 s to 15 sm/s²(C) 0 s to 20 sm/s²(ii) Find the instantaneous accelerations at the following times.(a) 2.0 sm/s²(b) 10 sm/s²(c) 18 sm/s²

Answer 1

(i)

(a) 0 m/s²

(b) 1.6 m/s²

(c) 0.8 m/s²

(ii)

(a) 0 m/s²

(b) 1.6 m/s²

(c) 0 m/s²

Explanation:

(i)

We have that the acceleration is given as follows:

`a=(\Delta v)/(\Delta t)`

(a) 0 s to 5 s:

`a=(v_5-v_0)/(5-0)=(-8-(-8))/(5-0)=(-8+8)/(5)=(0)/(5)=0\text{ }(m)/(s^2)`

(b) 5 s to 15 s:

`a=(v_(15)-v_5)/(15-5)=(8-(-8))/(15-5)=(8+8)/(10)=(16)/(10)=1.6\text{ }(m)/(s^2)`

(c) 0 s to 20 s:

`a=(v_(20)-v_0)/(20-0)=(8-(-8))/(20-0)=(8+8)/(20)=(16)/(20)=0.8\text{ }(m)/(s^2)`

(ii)

The instantaneous acceleration is given as follows:

`a=\frac{dv}{\text{ dt}}\text{ (slope of graph)}`

(a) 2 sec (No slope so acceleration at 2 sec is 0 m/s²). We confirm it as follows:

`a=(v_2-v_0)/(2-0)=(-8-(-8))/(2-0)=(-8+8)/(2)=(0)/(2)=0\text{ }(m)/(s^2)`

(b) 10 sec

The slope is:

`a=(8-(-8))/(15-5)=(8+8)/(10)=(16)/(10)=1.6\text{ }(m)/(s^2)`

(c) 18 sec: No slope so acceleration at 18 sec is 0 m/s²