Question

A car going 50km/h can stop over 15m. On the same road, in what distance can the same car be stopped when its speed is 70km/h?

Answer 1

the car can stop due to the work done by the friction forces. Now the magnitude R, of such friction forces are given by the subsequent fromula:
\[\Large \frac{1}{2}m{v^2} = Rd\]
where m is the mass of the car and d= 15 meters, v is the velocity of the car

Answer 2

so, solving that equation for R, we get:
\[\Large R = \frac{{m{v^2}}}{{2d}}\]

Answer 3

now, when the speed of the car is v_1=70 km/h, then the subsequent equation holds:
\[\Large \frac{1}{2}mv_1^2 = R{d_1}\]
where d_1 is the new distance, and we have to compute it.
Substituting for R, we get:
\[\Large \frac{1}{2}mv_1^2 = \frac{{m{v^2}}}{{2d}}{d_1}\]

Answer 4

finally, after a simplification, we can write:
\[\Large {d_1} = d{\left( {\frac{{{v_1}}}{v}} \right)^2}\]
where:
\[\Large {v_1} = 70\;Km/h\]

Answer 5

please substitute your data into that formula and you will get the requested distance

Answer 6

There is a simple formula for an object in acceleration/ retardation:

v^2=u^2+2aS

Where '
v=final velocity ( 0 here, as it is stopping)
u=initial velocity (50kmph= (50*5/18 )m/s here as that was the initial 'speed')
a= the acceleration that is acting on the vehicle (unknown, we'll find it out)
S= distance over which the acceleration works (15m)

Putting these values:
0^2=(50*5/18)^2+2*a*15
We get a=-6.43004 (m/s^2) This is a constant as the breaks of the car shall provide constant acceleration.

For the next case, we have v=0, u=70*5/18 m/s, a=-6.4301 m/s^2, s=unknown
Putting values:

0^2=(70*5/18)^2+2*-6.4301*S

S=29.3997 m