# Couple Physics Questions?

+1 vote

A subway train starting from rest leaves a
station with a constant acceleration. At the
end of 9.34 s, it is moving at 15.0374 m/s.
What is the train’s displacement in the ﬁrst
5.99628 s of motion?

A car is moving at a constant speed of 15 m/s
when the driver presses down on the gas pedal
and accelerates for 12 s with an acceleration
of 1.8 m/s^2
What is the average speed of the car during
the period?

Hang time means the time the player is oﬀ the
ground.
A basketball player achieves a hang time of
0.823 s in dunking the ball.
What vertical height will he attain? The
acceleration of gravity is 9.8 m/s^2

Im not sure how to do these, can someone explain? Thanks!!!
asked Jan 21, 2013 in PHYSICS

+1 vote
a)

Total time,T=9.34 s

Initial speed,u=0

Final speed,v=15.0374 m/s

---------------------------------

Using fundamental motion equation

v=u+aT

a=(v-u)/T

=(15.0374-0)/9.34

=1.61 m/s^2

Displacement at 5.99628 s

s=ut+(1/2)at^2

=0+(1/2)*(1.61)*(5.99628)^2

=28.944 m

-----------------------------

2)

Initial speed,u=15 m/s

Time period,t= 12 s

Accelaration,a=1.8 m/s^2

v=u+at

v=15+(1.8*12)

=36.6 m/s

Average speed,vavg=(v+u)/2

=(36.6+15)/2

=25.8 m/s

--------------------------------------------

3)

Vertical height,h=(1/2)gt^2

=(1/2)*9.81*0.823^2

=3.322 m