Question

A jogger accelerates from rest to 3.0 m/s in 2.0 s. A car accelerates from 38.0 to 41.0 m/s also in 2.0 s. (a) Find the acceleration (magnitude only) of the jogger, (b) Determine the acceleration (magnitude only) of the car. (c) Does the car travel farther than the jogger d?

Answer 1

Answer:

A- $a_{j} =1.5$$\frac{m}{s^{2} }$

B- $a_{c} =1.5$$\frac{m}{s^{2} }$

C- Yes, it does. The car travels 76.0m further than the jogger.

Explanation:

To calculate acceleration we have to use the following equation of motion:

$v=v_{o} +at$

$v:$Velocity

$v_{0}:$Starting velocity

$a:$Acceleration

$t:$Time

So now, we get acceleration as:

$a=\frac{v-v_{0} }{t}$

A- The jogger accelerates from rest, this means $v_{0} =0.0\frac{m}{s}$, to $v=3.0\frac{m}{s}$ in $t=2.0s$

Replacing the data in the acceleration fórmula:

$a_{j}=\frac{v-v_{0} }{t}$

$a_{j}=\frac{3.0\frac{m}{s} -0\frac{m}{s}}{2.0s}$

$a_{j}=1.5\frac{m}{s^{2} }$

B- The car accelerates from $v_{0} =38.0\frac{m}{s}$ to $v=41.0\frac{m}{s}$ in $t=2.0s$

Using the same formula from A:

$a_{c}=\frac{v-v_{0} }{t}$

$a_{c}=\frac{41\frac{m}{s}-38\frac{m}{s}}{2.0s}$

$a_{c}=1.5\frac{m}{s^{2} }$

C-Now we have to calculate distance to know if the car travels further than the jogger, we need to use the following formula:

$x=v_{0}t+\frac{1}{2} at^{2}$,

$x:$Distance.

For the jogger,

$x_{j}=v_{0}t +\frac{1}{2} at^{2}$

$x_{j} =0.0\frac{m}{s}.(0.0s)+\frac{1}{2} (1.5\frac{m}{s^{2} } )(2.0s)^{2}$

$x_{j}=3.0m$

For the car,

$x_{c}=v_{0} +\frac{1}{2} at^{2}$

$x_{c} =38\frac{m}{s} +\frac{1}{2} (1.5\frac{m}{s^{2}})(2s)^{2}$

$x_{c} =79.0m$

Then, $x_{c} -x_{j}=79.0m-3.0m=76.0m$

The car travels 76.0m further than the jogger.