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
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.
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
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
To solve these questions, we will use the equations of motion and the principles of Newtonian mechanics.
(a) The final speed of the car:
Since the car is initially moving at a constant speed and then accelerates at a constant rate, we can use the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
Given:
Initial velocity (u) = 30.0 m/s
Acceleration (a) = 3.00 m/s^2
Time (t) = 10.0 s
Using the equation, v = u + at, we can substitute the values and solve for v:
v = 30.0 m/s + (3.00 m/s^2)(10.0 s)
v = 30.0 m/s + 30.0 m/s
v = 60.0 m/s
Therefore, the final speed of the car is 60.0 m/s.
(b) The distance traveled during the acceleration:
To determine the distance traveled during the acceleration, we can use the equation of motion: s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time taken.
Given:
Initial velocity (u) = 30.0 m/s
Acceleration (a) = 3.00 m/s^2
Time (t) = 10.0 s
Using the equation, s = ut + (1/2)at^2, we can substitute the values and solve for s:
s = (30.0 m/s)(10.0 s) + (1/2)(3.00 m/s^2)(10.0 s)^2
s = 300.0 m + (1/2)(3.00 m/s^2)(100.0 s^2)
s = 300.0 m + 150.0 m
s = 450.0 m
Therefore, the car travels a distance of 450.0 meters during the acceleration.
(c) The overall force acting on the car during the acceleration:
According to Newton's second law of motion, force (F) is equal to the product of mass (m) and acceleration (a): F = ma.
Given:
Mass (m) = 1000 kg
Acceleration (a) = 3.00 m/s^2
Using the equation, F = ma, we can substitute the values and solve for F:
F = (1000 kg)(3.00 m/s^2)
F = 3000 N
Therefore, the overall force acting on the car during the acceleration is 3000 Newtons.
(d) The initial momentum of the car:
Momentum (p) is equal to the product of mass (m) and velocity (v): p = mv.
Given:
Mass (m) = 1000 kg
Initial velocity (u) = 30.0 m/s
Using the equation, p = mv, we can substitute the values and solve for p:
p = (1000 kg)(30.0 m/s)
p = 30000 kg*m/s
Therefore, the initial momentum of the car is 30000 kg*m/s.
(e) The initial kinetic energy of the car:
Kinetic energy (KE) is given by the equation: KE = (1/2)mv^2, where m is mass and v is velocity.
Given:
Mass (m) = 1000 kg
Initial velocity (u) = 30.0 m/s
Using the equation, KE = (1/2)mv^2, we can substitute the values and solve for KE:
KE = (1/2)(1000 kg)(30.0 m/s)^2
KE = (1/2)(1000 kg)(900.0 m^2/s^2)
KE = 450000 J
Therefore, the initial kinetic energy of the car is 450000 Joules.