A car with a mass of 1264 kg is coasting in neutral on a straight,level road. It slows down, and its speed as a function of time is given by the equation:
v(t) = a − bt + ct2
Constant Value Units
a 26.8 m/s
b 0.310 m/s2
c 2.10*10^-3 m/s2
At a time of 36.0 s the speed, as given by the above equation, is 18.36 m/s. Calculate the power which the engine must deliver (to compensate for air resistance and rolling resistance) in order to maintain that speed.
To calculate the power that the engine must deliver to maintain the speed, we need to calculate the net force acting on the car and then use the equation for power.
First, let's find the acceleration of the car. Taking the derivative of the given equation for velocity with respect to time will give us the acceleration:
a(t) = -b + 2ct
Substituting the given values of b and c into the equation:
a(t) = -0.310 + 2(2.10*10^-3)t
= -0.310 + 0.0042t
Now, we can find the time at which the speed is 18.36 m/s by solving the equation v(t) = 18.36:
18.36 = 26.8 - (0.310)t + (2.10*10^-3)(t^2)
Since this is a quadratic equation, we can rearrange it to standard form:
(2.10*10^-3)t^2 - (0.310)t + (26.8 - 18.36) = 0
(2.10*10^-3)t^2 - (0.310)t + 8.44 = 0
We can use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values of a, b, and c:
t = (0.310 ± √((-0.310)^2 - 4(2.10*10^-3)(8.44))) / (2(2.10*10^-3))
After solving this equation, we find two possible values for t: t = 9.30 s or t = 33.5 s.
However, since we are given that the time is 36.0 s, the car does not have a speed of 18.36 m/s at that time. Therefore, the information given is not consistent.
a=dv/dt= -b+2ct =-0.31+2.1•10⁻³•36=
=-0.2344 m/s²
P=Fv=ma•v = 1264•0.2344•18.36 = 5439.7 W