Question

A 7,500 kg truck lost its breaks as it crested a 500 m hill. The downward angle of the hill is 30 degrees clockwise from the direction of motion. Before the driver realized the breaks were gone she was going 24 m/s. This is a narrow roadway and the next hill (where she can park safely) is 1200 m high at 35 degrees counter-clockwise from the direction of motion. Ignore air resistance and friction.

a) What is her maximum speed during this part of her trip?

b) How far up the hill will she go?

c) Will she reach the top of the next hill, if not what intial velocity would get her there?

Answer 1

a) vertical height of the first hill, h = d1*sin(30)

= 500*sin(30)

= 250 m

maximum speed at the bootm of the first hill,

v_max = sqrt(vo^2 + 2*g*h1)

= sqrt(24^2 + 2*9.8*250)

= 74 m/s <<<<<<<<<<<------------------Answer

b) vertical height reached on second hill,

h2 = vmax^2/(2*g)

= 74^2/(2*9.8)

= 279.4 m

distance travelled along the second hill, d2 = h2/sin(theta2)

= 279.4/sin(35)

= 487 m <<<<<<<<<<<------------------Answer

c) h2_max = 1200*sin(35)

= 688 m

so, (1/2)*m*v^2 + m*g*h1 = m*g*h2_max

v = sqrt(2*g*(h2max - h1))

= sqrt(2*9.8*(688 - 250))

= 92.7 m/s <<<<<<<<<<<------------------Answer