3.3 Average and instantaneous acceleration  (Page 4/7)

Solution

1. $a(t)=\frac{dv(t)}{dt}=20-10t{\text{m/s}}^2$
2. $v(1\text{s})=15\text{m/s}$ , $v(2\text{s})=20\text{m/s}$ , $v(3\text{s})=15\text{m/s}$ , $v(5\text{s})=\mathrm{-25}\text{m/s}$
3. $a(1\text{s})=10{\text{m/s}}^2$ , $a(2\text{s})=0{\text{m/s}}^2$ , $a(3\text{s})=\mathrm{-10}{\text{m/s}}^2$ , $a(5\text{s})=\mathrm{-30}{\text{m/s}}^2$
4. At t = 1 s, velocity $v(1\text{s)}=15\text{m/s}$ is positive and acceleration is positive, so both velocity and acceleration are in the same direction. The particle is moving faster.

At t = 2 s, velocity has increased to $v(2\text{s)}=20\text{m/s}$ , where it is maximum, which corresponds to the time when the acceleration is zero. We see that the maximum velocity occurs when the slope of the velocity function is zero, which is just the zero of the acceleration function.

At t = 3 s, velocity is $v(3\text{s)}=15\text{m/s}$ and acceleration is negative. The particle has reduced its velocity and the acceleration vector is negative. The particle is slowing down.

At t = 5 s, velocity is $v(5\text{s)}=\mathrm{-25}\text{m/s}$ and acceleration is increasingly negative. Between the times t = 3 s and t = 5 s the particle has decreased its velocity to zero and then become negative, thus reversing its direction. The particle is now speeding up again, but in the opposite direction.

We can see these results graphically in [link] .

Significance

Getting a feel for acceleration

You are probably used to experiencing acceleration when you step into an elevator, or step on the gas pedal in your car. However, acceleration is happening to many other objects in our universe with which we don’t have direct contact. [link] presents the acceleration of various objects. We can see the magnitudes of the accelerations extend over many orders of magnitude.