Question

1.     A driver of an ambulance going 70 km/h suddenly sees another car 35.0 m ahead. It takes the driver 0.62 s before he applies the brakes. Once he does begin to brake, he reduces speed at a rate of 13.0 m/s².

 

1.1  Determine if the driver will hit the car or not.                                               (6)

 

What would be the maximum speed at which the ambulance could travel and NOT hit the car 35.0m ahead?             

Answer 1

$a=\frac{v-v_0}{t}=13$ m/s²

Time of breaking:

$t=\frac{v-v_0}{a}=70/(3.6\cdot13)=1.5\ s$

Breaking distance:

$d=70\cdot0.62/3.6+v_0t-at^2/2$

$d=70\cdot0.62/3.6+70\cdot1.5/3.6-13\cdot1.5^2/2=26.6\ m$

Since distance d < 35 m, the driver will not hit the car.

For maximum speed:

$d=35=v_{max}\cdot0.62+v_{max}t-at^2/2$

$d=35=v_{max}\cdot0.62+v_{max}^2/a-v^2_{max}/(2a)=v_{max}\cdot0.62+v^2_{max}/(2a)$

$v^2_{max}+16.12v_{max}-910=0$

$v_{max}=\frac{-16.12+\sqrt{16.12^2+4\cdot910}}{2}=23.16\ m/s=83.4\ km/h$