Question

After an initial race George determines that his car loses 35 percent of its acceleration due to air resistance travelling at 38 m/s on flat ground. Assuming that his car travels with a constant acceleration, calculate the maximum speed (Vm), in meters per second, his car can reach on flat ground?\

Answer 1

Answer:

64.2 m/s

Explanation:

We are given that

Speed ,v=38 m/s

We have to find the maximum speed when his car reach on flat ground.

Using dimensional analysis

$F_{res}\propto v^2$

If 35% acceleration reduced by F(res) at 38 m/s

Then, 100% acceleration  can be reduced  by F(res) at v' m/s

$\frac{F_1}{F_2}=\frac{v^2}{v'^2}$

$v'^2=\frac{F_2}{F_1}v^2$

$v'=v\sqrt{\frac{F_2}{F_1}}$

Substitute the values

$v'=38\times \sqrt{\frac{100}{35}}$

$v'=64.2 m/s$

Hence, the maximum speed when his car can reach on flat ground=64.2 m/s

Answer 2

The maximum velocity of the car at constant acceleration on the road will be 62.2 m/s.

What is the relation between resistance and speed?

The air resistance is directly proportional to the square of the velocity of an object.

R ∝ v²

The speed of the car was reduced by 35 % at 38 m/s.

So, the speed of the car was reduced by 100% at v' m/s.

The relationship can be given by,

$\dfrac {R_1}{R_2} = \dfrac {v^2}{v'^2}$

Put the values in the formula and calculate for $v'$,

$v' = \sqrt {\dfrac {R_2 v^2}{R_1 }}$

$v' = \sqrt {\dfrac { 100\times 38 ^2}{35 }}\\\\v' = 62.2 \rm \ m/s$

Therefore, the maximum velocity of the car at constant acceleration on the road will be 62.2 m/s.

Learn more about resistance and speed?

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