Find the time

Question

A professional race-car driver buys a car
that can accelerate at 5.5 m/s²
. The racer
decides to race against another driver in a
souped-up stock car. Both start from rest,
but the stock-car driver leaves 1.5 s before the
driver of the sports car. The stock car moves
with a constant acceleration of +4.4 m/s²
.
Find the time it takes the sports-car
driver to overtake the stock-car driver.

Answer 1

THE PHYSICS EQN IS JUST  X = ACC*T²/2  + V*T + X0

 

FOR BOTH CARS, THEY HAVE NO INITIAL SPEED V1  OR DISPLACEMENT X0

 

SO X = ACC*T²/2 

 

CAR A HAS ACC = 5.5  BUT LEAVES 1.5 SEC AFTER CAR B

CAR B HAS ACC = 4.4

 

X_B = 4.4*T²/2      FOR CAR B

X_A = 5.5*T_A²/2    FOR CAR A     (USE T_A = T-1.5)

X_A = 5.5*(T-1.5)²/2

 

EXPANDING X_A = 5.5 * (T² -3T + 9/4) / 2

 

NOW JUST SET X_A = X_B   AND SOLVE FOR T (TIME WHICH CAR B STARTS)

 

(T=TIME THEY WILL BE EQUAL DIST FROM THE START)

 

4.4*T²/ 2 =  5.5*(T² -3T + 9/4) / 2     MULT OUT THE 2 ON EACH SIDE

 

4.4*T² = 5.5*T² - 16.5T + 12.375       NOW SUBT 4.4*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=1.5 WHEN CAR A STARTS,

 

SO WE USE THE (+) PART   WHICH GIVES T = 14.2082 SEC  (CAR B)

 

T = TIME THE FIRST CAR B STARTED, CAR A STARTED 1.5 SEC LATER

CAR A TIME T_A =  T - 1.5 OR  (14.2082 - 1.5) SEC  OR T_A = 12.7082  (CAR A)

 

AS A CHECK,

CAR B (LEAVES FIRST AT T=0)  

X_B = 4.4*(14.2082)² / 2   OR   X_B = 444.1207 METERS

 

CAR A (LEAVES AT T = 1.5 SEC)

SO AT T = 14.2082 SEC, IT HAS DRIVEN FOR T_A = 12.7082 SEC)

 

X_A = 5.5*(12.7082)² / 2  OR  X_A = 444.1207 METERS    SAME DIST AS B