Question

A car initially traveling at 32.4 m/s undergoes a constant negative acceleration of magnitude 1.70 m/s2 after its brakes are applied. (a) How many revolutions does each tire make before the car comes to a stop, assuming the car does not skid and the tires have radii of 0.330 m?

Answer 1

The car tire makes 149 revolutions before coming to stop.

Explanation:

a) We have equation of motion v² = u² + 2as

Initial velocity, u = 32.4 m/s  

Acceleration, a = -1.70 m/s²  

Final velocity, v = 0 m/s  

Substituting  

v² = u² + 2as

0² = 32.4² + 2 x -1.70 x s

s = 308.75 m  

Distance traveled before stopping is 308.75 m

Radius of tire = 0.330 m

Circumference of tire = 2πx 0.33 = 2.07 m

1 revolution = 2.07 m

$\texttt{Number of revolutions = }\frac{308.75}{2.07}=148.91$

The car tire makes 149 revolutions before coming to stop.

Answer 2

Before coming to the stop, the car times makes 149 revolutions.

Given values:

• Initial velocity, $u = 32.4 \ m/s$
• Final velocity, $v = 0 \ m/s$
• Acceleration, $a = -1.70 \ m/s^2$
• Radius of tire, $r = 0.330 \ m$

By using the equation of motion,

→ $v^2=u^2+2as$

By putting the values,

  $0^2=32.4^2+2(-1.70) s$

    $s = 308.75 \ m$

As we know,

• 1 revolution = 2.07 m

hence,

The number of revolutions will be:

= $\frac{308.75}{2.07}$

= $148.91$

or,

= $149$

Thus the answer above is right.

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