Question

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

Answer 1

Answer:

Initial velocity=9 m/s

Acceleration=0.767 m/s^2

Answer 2

Answer:

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.