Question

A 1300 kg car is to accelerate from rest to a speed of 30.0m/s in a time of 12 s as it climbs a 15 degrees hill. Assuming uniform acceleration, what minimum horsepower is needed to accelerate the car in this way?​

Answer 1

Given :

• Mass of the car = 1300 kg

• Velocity of car = 30 m/s

• Time = 12 s

• Angle of Projection = 15°

To find :

Horsepower needed to accelerate the car.

Solution :

First let us find the acceleration Produced by car.

We know the first Equation of Motion i.e,

$\bf{v = u + at}$

Where :

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time Taken

[Note : Here , the initial velocity of the car will be zero, u = 0 , since the car is starting from rest.]

Using the first Equation of Motion and substituting the values in it, we get :

$:\implies \bf{v = u + at}$

$:\implies \bf{30 = 0 + a \times 12}$

$:\implies \bf{30 = 12a}$

$:\implies \bf{\dfrac{30}{12} = a}$

$:\implies \bf{2.5 = a}$

$\boxed{\therefore \bf{a = 2.5\:ms^{-2}}}$

Hence, the acceleration produced by the car is 2.5 m/s².

Now let's find the distance covered by the car :

We know the second Equation of Motion i.e,

$\bf{S = ut + \dfrac{1}{2}at^{2}}$

Where :

• S = Distance
• t = Time Taken
• a = Acceleration
• u = Initial Velocity

Using the second Equation of Motion and substituting the values in it, we get :

$:\implies \bf{S = ut + \dfrac{1}{2}at^{2}}$

$:\implies \bf{S = 0 \times t + \dfrac{1}{2}at^{2}}$

$:\implies \bf{S = \dfrac{1}{2}at^{2}}$

$:\implies \bf{S = \dfrac{1}{2} \times 2.5 \times 12^{2}}$

$:\implies \bf{S = \dfrac{1}{2} \times 2.5 \times 144}$

$:\implies \bf{S = 2.5 \times 72}$

$:\implies \bf{S = 180}$

$\boxed{\therefore \bf{S = 180\:m}}$

Hence the distance traveled by the car is 180 m.

Now , to find the power of the car.

We know that ,

$\bf{P = \dfrac{W}{t}}$

Where :

• P = Power
• W = Work Done
• t = Time Taken

But we know that ,

$\boxed{\bf{W = Fs}}$⠀⠀⠀⠀⠀⠀Eq.(i)

Where :

• W = Work Done
• F = Force
• s = Displacement

We also know that :

$\bf{F = ma}$⠀⠀⠀⠀⠀⠀⠀Eq.(ii)

Where :

• F = Force
• m = Mass
• a = Acceleration

Now , by putting the value of Work Done (W) from Eq.(i) in the equation for power , we get :

$:\implies \bf{P = \dfrac{Fs}{t}}$

By putting the value of Force (F) from Eq.(ii) in the above equation , we get :

$:\implies \bf{P = \dfrac{ma\:s}{t}}$

Now by substituting the given values in the equation, we get :

$:\implies \bf{P = \dfrac{1300 \times 2.5 \times 180}{12}}$

$:\implies \bf{P = \dfrac{1300 \times 2.5 \times 180}{12}}$

$:\implies \bf{P = \dfrac{585000}{12}}$

$:\implies \bf{P = 48750}$

$\boxed{\therefore \bf{P = 48750\:W}}$

Hence, the Power of the car is 48750 W

Now , let's convert it in Horse Power.

We know that ,

1 H·P = 746 W

Hence,

==> 48750 W = (48750/746) H·P

==> 48750 W = 65.34 (Approx.) H·P

Hence the horse power of the car is 65.34 H·P

Answer 2

Answer:

The minimum horse power needed is 232.02H.p

Explanation:

Given that,

Mass of car(m)=1300kg

Initial velocity(u)=0

Final velocity(v)=30m/s

Time taken(t)=12sec

Inclination=15°

We need to find acceleration in this case,

$v = u + at \\ 30 = 0 + a(12) = > a = 2.5ms {}^{ - 2}$

Also we need to find the distance covered by car

$s = ut + \frac{1}{2} at {}^{2} \\ s = 0(t) + \frac{1}{2} (2.5)12 {}^{2} = 180m$

As acceleration and distance are known,we can find the energy of car

$E = \frac{1}{2} mv {}^{2} + mgh \\ = \frac{1}{2} mv {}^{2} + mgs(sin15 {}^{0} ) \\ = \frac{1}{2} \times 1300 \times 30 {}^{2} + 1300 \times 9.81 \times 180 \times 0.65 \\ = 2077101J$

Now power can be found out using relation,

$power = \frac{energy}{time} = \frac{2077101}{12} = 173091.75W$

We know that 1 Horse power(H.P)=746W

Therefore minimum horsepower needed is

$\frac{173091.75}{746 } = 232.02HP$

#SPJ2