Physics, 20.09.2019 01:00 shilohtito

Ajogger 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 during the 2.0 s? if so, how much farther?

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Answer from: bob5266

a) The acceleration of the jogger is 1.5 m/s²

b) the acceleration of the car is also 1.5 m/s²

c) Yes, the car travels 76 m farther than the jogger.

Explanation:

a) The acceleration of an object is the variation of its velocity over time:

a = final velocity - initial velocity / time

for the jogger:

a = 3.0 m/s - 0 m/s / 2.0 s = 1.5 m/s ²

b) For the car:

a = 41.0 m/s - 38.0 m/s / 2.0 s = 1.5 m/s²

c) Let´s see the position of the car after 2 seconds.

The equation for the position of an accelerated object moving in a straight line is:

x = x0 + v0* t +1/2 * a * t²

Where:

x = position of the car at time "t"

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Let´s consider x0 = 0 because the origin of the reference system is located where the car starts accelerating. Then:

x = 38,0 m/s * 2 s + 1/2 * 1.5 m/s ² * (2.0 s)²

x = 79 m

In the same way, we can calculate the position of the jogger:

x = 0 m/s * t + 1/2 * 1.5 m/s ² * (2.0 s)²

x = 3 m

The car travels 79 m - 3 m = 76 m farther than the jogger

Answer from: smagallanes

(a) The acceleration of the jogger : $a_{j}=1.5 \mathrm{~m} \cdot \mathrm{s}^{-2}$

(b) The acceleration of the car : $a_{c}=1.5 \mathrm{~m} \cdot \mathrm{s}^{-2}$

Explanation:

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 $2.0 \ s$

Step 1:

Using the following equation of motion:

$v=v_{0}+a t$

we can get the acceleration as:

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

(a) If the jogger start from rest $v_{0}=0$ to a speed of $v=3 \mathrm{~m} \cdot \mathrm{s}^{-1}$ in time interval of $t=2 \mathrm{~s}$ , the acceleration is:

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

$=\frac{\left(3 \mathrm{~m} \cdot \mathrm{s}^{-1}\right)-(0)}{(2)}$

$&=1.5 \mathrm{~m} \cdot \mathrm{s}^{-2} \end{aligned}$$

Step 2:

(b) If the car start from the rest $v_{0}=38 \mathrm{~m} \cdot \mathrm{s}^{-1}$ to a speed of $v=41 \mathrm{~m} \cdot \mathrm{s}^{-1}$ in a time interval of $t=2 \mathrm{~s}$ , the acceleration is:

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

$=\frac{\left(41 \mathrm{~m} \cdot \mathrm{s}^{-1}\right)-\left(38 \mathrm{~m} \cdot \mathrm{s}^{-1}\right)}{(2)}$

$=1.5 \mathrm{~m} \cdot \mathrm{s}^{-2}$

Step 3:

Using the following equation we can calculate the distance:

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

From the jogger,

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

$&=0+\frac{1}{2}\left(1.5 \mathrm{~m} \cdot \mathrm{s}^{-2}\right)(2 \mathrm{~s})^{2} \\$

$&=3 \mathrm{~m} \end{aligned}$$

$x_{j}=4.5 \mathrm{~m}$

Step 4:

For the car,

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

$&=\left(38 \mathrm{~m} \cdot \mathrm{s}^{-1}\right)(2)+\frac{1}{2}\left(1.5 \mathrm{~m} \cdot \mathrm{s}^{-2}\right)(2 \mathrm{~s})^{2} \\$

$&=79 \mathrm{~m} \end{aligned}$$

$x_{c}=79 \mathrm{~m}$

Therefore, the car travles $x_{c}-x_{j}=76 \mathrm{~m}$ further than the jogger.

To learn more about acceleration, refer:

link

Answer from: Quest

i think its c if i'm wrong

Answer from: Quest

It equals just about 6,300 km per hour each day an week

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Ajogger 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...