How long will it take a 2.3"x10^3 kg truck to go from 22.2 m/s to a complete stop if acted on by a force of -1.26x10^4 N.What wo

Question

4 years ago

6

uld be it's stopping distance?

1 answer:

Greeley [361] · Answer 1 · 2021-11-07T03:03:16+03:00

The stopping distance is 45.0 m

Explanation:

First of all, we find the acceleration of the truck, by using Newton's second law:

$F=ma$

where

$F=-1.26\cdot 10^4 N$ is the net force on the truck

$m=2.3\cdot 10^3 kg$ is the mass of the truck

a is its acceleration

Solving for a,

$a=\frac{F}{m}=\frac{-1.26\cdot 10^4}{2.3\cdot 10^3}=-5.48 m/s^2$

where the negative sign means the acceleration is opposite to the direction of motion.

Now, since the motion of the truck is at constant acceleration, we can apply the following suvat equation:

$v^2-u^2=2as$

where

v = 0 is the final velocity of the truck

u = 22.2 m/s is the initial velocity

$a=-5.48 m/s^2$ is the acceleration

s is the stopping distance

And solving for s,

$s=\frac{v^2-u^2}{2a}=\frac{0-(22.2)^2}{2(-5.48)}=45.0 m$

Learn more about accelerated motion:

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