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 7.90 s to a final velocity of 2.00 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/s²) 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/s²

Answer 1

Given
the van moving with initial velocity v1 , final velocity is v2

   v1=? , v2 = 2.0 m/s

distance covered in time t = 7.90 s is , s = 40.0 m

a)What was its initial speed (in m/s), just before the brakes were applied?

m/s = v1

From work energy theorem the work done by the brakes makes the van, change in its kinetic energh

that is W = k2-k1

   F*s = k2-k1

   m*a*s = k2-k1

   m (v2-v1)/(t2-t1) *s = k2-k1

   m (v2-v1)/(t2-t1) *s = 0.5*m(v2^2-v1^2)

   2(v2-v1)/(t2-t1) *s = (v2^2-v1^2)
   2(v2-v1)/(t2-t1) *s = (v2+v1)(v2-v1)

   2*s/(dT) = (v2+v1)

   v1 = (2*s/dT)-v2

substituting the values

   v1 m/s = (2*40 m)/(7.9s) - (2m/s)

   v1 = 8.127 m/s

the initial velocity of the van is v1 = 8.127 m/s

b) From equations of motions

   v = u+at

   a = v-u/t

   a = (2m/s-8.127m/s)/(7.90s)

   a = -0.77557 m/s2
the acceleration (in m/s2) of the wan while the brakes were applied is a = -0.77557 m/s2