Question

Cheri has a new Dodge Charger with a Hemi engine and has challenged Vince, who owns a tuned VW GTI, to a race at a local track. Vince knows that Cheri’s Charger is rated to go from 0 to 60 mph in 5.3 s, whereas his VW needs 7 s. Vince asks for a head start and Cheri agrees to give him exactly 1 s. (a) How far down the track is Vince before Cheri gets to start the race? (b) At what time does Cheri catch Vince? (c) How far away from the start are they when this happens? (Assume constant acceleration for each car during the race.)

Answer 1

Answer:

a) 1.915 s

b) 7 s

c) 122.5 m

Explanation:

1 mph = 0.44704 m/s

60 mph = (0.44704 * 60) m/s = 26.8 m/s

a) The VW acceleration = $a_1$ = (26.8 m/s - 0 m/s) / 7 s =  3.83 m/s²

Cheri's acceleration = $a_2$ = (26.8 m/s - 0 m/s) / 5.3 s = 5 m/s²

The distance traveled is given as:

$S=ut+\frac{1}{2}at^2\\ \\For\ Vince\ VW\ which\ has\ traveled\ for\ 1\ s:\\\\S=0(1)+\frac{1}{2}(3.83)(1)^2=1.915\ m$

b) The distance traveled by Vince VW with a 1s head start is given as:

$S_1=ut+\frac{1}{2}a_1(t+1)^2\\ \\S_1=\frac{1}{2}(3.83)(t+1)^2\\\\S_1=1.915(t+1)^2\\\\S_1=1.915(t^2+2t+1)\\\\S_1=1.915t^2+3.83t+1.915$

The distance traveled by Cheri is given as:

$S_2=ut+\frac{1}{2}a_2(t)^2\\ \\S_2=\frac{1}{2}(5)(t)^2\\\\S_2=2.5t^2\\\\S_1=S_2\\\\2.5t^2=1.915t^2+3.83t+1.915\\\\0.585t^2-3.83t-1.915=0\\\\t=7s\ or\ t=-0.5\ s\\\\t=7s$

c) $S_1=2.5t^2\\\\S_1=2.5(7)^2=122.5\ m$