Question
A car of mass 1000kg is travelling along a straight flat road at a constant speed of 30.0ms^-1. It then accelerates with constant acceleration of 3.00ms^-2 for 10.0s, after which it continues to travels at a constant speed. Determine:
(a) The final speed of the car
(b) The distance did the car travelled during the acceleration
(c) The overall force acting on the car during the acceleration
(d) The initial momentum of the car
(e) The initial kinetic energy of the car
(a) The final speed of the car
(b) The distance did the car travelled during the acceleration
(c) The overall force acting on the car during the acceleration
(d) The initial momentum of the car
(e) The initial kinetic energy of the car
Answers
Steve
just use your basic equations of motion:
v = Vo + at
s = Vo*t + 1/2 at^2
p = mv
KE = 1/2 mv^2
momback if you get stuck. And say where.
v = Vo + at
s = Vo*t + 1/2 at^2
p = mv
KE = 1/2 mv^2
momback if you get stuck. And say where.
Saphina AlMatary
for a I got:
30.0×10.0+0.5×(3.00+10.0)^2=384.5
so the final speed is 384.5 m s???
From b to d its extremely difficult to make an equation for!
Please help me asap,
Kind Regards
30.0×10.0+0.5×(3.00+10.0)^2=384.5
so the final speed is 384.5 m s???
From b to d its extremely difficult to make an equation for!
Please help me asap,
Kind Regards
Debele Bedada
Solution.
a)v=u+at
=30+3*10
=60m/s
b)s=vt+1/2at^2
=30*10+0.5*3*10*10
=450m
c)F=ma
=1000kg*3m/s^2
=3000N
d)P=mv
=1000kg*30m/s
=30000unit
e)KE=1/2mv^2
=0.5*1000kg*30^2
=40000J
a)v=u+at
=30+3*10
=60m/s
b)s=vt+1/2at^2
=30*10+0.5*3*10*10
=450m
c)F=ma
=1000kg*3m/s^2
=3000N
d)P=mv
=1000kg*30m/s
=30000unit
e)KE=1/2mv^2
=0.5*1000kg*30^2
=40000J