Physics, 14.09.2019 03:30 lolyan555

The brakes are applied to a moving van, causing it to uniformly slow down. while slowing, it moves a distance of 40.0 m in 7.70 s to a final velocity of 1.80 m/s, at which point the brakes are released. (a) what was its initial speed (in m/s), just before the brakes were applied? m/s (b) what was its acceleration (in m/s^2) while the brakes were applied? (assume the initial direction of motion is the positive direction. indicate the direction with the sign of your answer.) m/s^2

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Answer from: diawia

When the brakes are applied to a moving van, it travels a distance of 40.0 m in 7.70 s with a final velocity of 1.80 m/s.

a) The initial speed of the van just before the brakes were applied was 8.59 m/s.

b) The acceleration of the van while the brakes were applied was -0.88 m/s².

a) The initial speed of the van can be calculated as follows:

$v_{f} = v_{i} + at$

Where:

$v_{f}$ : is the final velocity = 1.80 m/s

$v_{i}$ : is the initial velocity =?

a: is the acceleration

t: is the time = 7.70 s

By solving the above equation for a we have:

$a = \frac{v_{f} - v_{i}}{t}$ (1)

Now, we need to use other kinematic equation to find the initial velocity.

$v_{i}^{2} = v_{f}^{2} - 2ad$ (2)

By entering equation (1) into (2) we have:

$v_{i}^{2} = v_{f}^{2} - 2d(\frac{v_{f} - v_{i}}{t}) = (1.80 m/s)^{2} - 2*40.0 m(\frac{1.80 m/s - v_{i}}{7.70 s})$

After solving the above equation for $v_{i}$ we get:

$v_{i} = 8.59 m/s$

Hence, the initial velocity is 8.59 m/s.

b) The acceleration can be calculated with equation (1):

$a = \frac{v_{f} - v_{i}}{t} = \frac{1.80 m/s - 8.59 m/s}{7.70 s} = -0.88 m/s^{2}$

Then, the acceleration is -0.88 m/s². The minus sign is because the van is decelerating.

You can find more about acceleration here: link

I hope it helps you!

Answer from: quanharris2k19

a)8.59 m/s

b)-0.8818 m/s²

Explanation:

a) Given the van moved 40 m in 7.70 seconds to a final velocity of 1.80 m/s

Apply the equation for motion;

$d=(\frac{V_i+V_f}{2} )*t$

where

t=time the object moved

d=displacement of the object

Vi=initial velocity

Vf=final velocity

Given

t=7.70s

Vf=1.80 m/s

d=40m

Vi=?

Substitute values in equation

$40=(\frac{V_i+1.80}{2} )7.70\\\\\\40=\frac{7.70V_i+13.86}{2} \\\\80=7.70V_i+13.86\\\\80-13.86=7.70V_i\\\\66.14=7.70V_i\\\\\frac{66.14}{7.70} =\frac{7.70V_i}{7.70} \\8.59=V_i$

b)Acceleration is the rate of change in velocity

Apply the formula

Vf=Vi+at

where;

Vf=final velocity of object

Vi=Initial velocity of the object

a=acceleration

t=time the object moved

Substitute values in equation

Given;

Vf=1.80 m/s

Vi=8.59 m/s

t=7.70 s

a=?

Vf=Vi+at

1.80=8.59+7.70a

1.80-8.59=7.70a

-6.79=7.70a

-6.79/7.70=7.70a/7.70

-0.8818=a

The van was slowing down.

The brakes are applied to a moving van, causing it to uniformly slow down. while slowing, it moves a

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grav force prop 1/r squared 2r all squared is 4rsquared 1/4rsquared quartered

Answer from: Quest

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The brakes are applied to a moving van, causing it to uniformly slow down. while slowing, it moves a...