A professional race-car driver buys a car that can accelerate at 6.4 m/s^2. The racer decides to race against another driver in a souped-up stock car. Both start from the rest, but the stock-car driver leaves 1.2 s before the driver of the sports car. The stock car moves with a constant acceleration of +3.3 m/s^2. Find the time it takes the sports-car driver to overtake the stock-car driver.
The answer is in units of s.
Answers
Answer:
9m/s² is the answer for the question
Answer:
Unit value is 1.2S.
Explanation:
The physics eqn is just
x = acc*t2/2 + v*t + x0
Displacement x0
So
x = acc*t2/2
Car a has acc = 6.4
But leaves 1.2 sec
after car b
car b has acc = 3.3
xb = 3.3*t2/2 for car b
xa = 6.4*ta2/2 for car a
(use ta = t-1.2)
xa = 6.4*(t-1.2)2/2
Expanding xa = 6.4* (t2 -3t + 9/4) / 2
now just set
xa = xb
and
Solve for t (time which car b starts)
3.3*t2/ 2 = 6.4*(t2 -3t + 9/4) / 2 mult out the 2 on each side
3.3*t2 = 6.4*t2 - 16.5t + 12.375
Now subt
3.3*t2
From each side
1.1*t2 - 16.5t + 12.375 = 0 a = 1.1
b = -16.5
c = 12.375
Solve by quadratic eqn of form ax2 + bx + c
t = -(-16.5) +/- sqrt( (-16.5)2 - 4*1.1*12.375) ) / (2*1.1)
t = (16.5 +/- 14.75805 ) / 2.2
Now, if we use the minus part, or
(16.5 - 14.758)/2.2
t= 0.792
This is not valid,
Since at
t=1.2s
When car a starts,
So the unit value is sec.