Question

A racing car reaches the straightaway on the race track going 140km/hr and accelerates at a rate of 3.5 m/s² for 6.1 seconds A) what is the car's initial velocity in m/s? B) what is the car's velocity at the end of the 6.2 seconds C) how far did the car travel in that time

Answer 1

Explanation:

c how far did the car travel in that time

Answer 2

Given,

Initial velocity, u=140 km/h

Acceleration, a=3.5  m/s²

Time, t=6.1 seconds.

To find,

A) what is the car's initial velocity in m/s?

B) what is the car's velocity at the end of the 6.2 seconds?

C) how far did the car travel in that time?

Solution:

A)To convert a speed given in km/h to m/s, we multiply it with $\frac{5}{18}$.

So, initial velocity in m/s will be:

$=140X\frac{5}{18} \\=\frac{700}{18} m/s\\=38.89 m/s.$

The initial velocity of the car in m/s is 38.89 m/s.

B)According the equation of motion,

$v=u+at\\v=38.89+3.5X6.1\\v=38.89+21.35\\v=60.24 m/s.$

The final velocity of the car after 6.1 seconds is 60.24 m/s.

C)According the equation of motion,

$s=ut+\frac{1}{2} at^{2} \\s=38.89X6.2+\frac{1}{2} X3.5X6.2^{2}\\s=241.118+\frac{1}{2} X3.5X6.2X6.2\\s=241.118+\frac{1}{2} X134.540\\s=241.118+67.27\\s=308.388 m\\s=308.4 m.$

The car travelled a distance of 308.4 metres in 6.2 seconds.

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