Suppose a rocket-propelled motorcycle is fi red from rest horizontally across a canyon 1.00 km wide. (a) What minimum constant acceleration in the x-direction must be provided by the engines so the cycle crosses safely if the opposite side is 0.750 km lower than the starting point? (b) At what speed does the motorcycle land if it maintains this constant horizontal component of acceleration? Neglect air drag, but remember that gravity is still acting in the negative y-direction.

Motion equation
x
=
a
t
2
2
x=
2
at
2



y
=
−
g
t
2
2
y=−
2
gt
2



Initial conditions
g
=
9.8
m
s
2
g=9.8
s
2

m


the cycle crosses safely if the...
1000
=
a
t
2
2
1000=
2
at
2



−
750
=
−
g
t
2
2
−750=−
2
gt
2



From this we have acceleration
a
=
1000
⋅
9.8
750
=
13.1
m
s
2
a=
750
1000⋅9.8

=13.1
s
2

m

Answer: it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.explanation:   we have equation of motion ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/sso it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.

Ithink it’s 98 because jajajjajansmsmjs