A Motorcycle Catches a Car! Physics Help Please?

Jerald Florence: Let M = distance travelled by the motorcycle when it catches up with the carM = 46(T2) + (1/2)(a)(T2)^2M = 46(T2) + (1/2)(8)(T2)^2M = 46(T2) + 4(T2)^2Let C = distance travelled by the car when the motorcycle catches up with itC = 46(T2)When the motorcycle catches up with the car, thenM = 59 + CTherefore,46(T2) + 4(T2)^2 = 59 + 46(T2)Since "46(T2)" appears on both sides of the equation, it will simply cancel out, hence the above equation is modified to4(T2)^2 = 59(T2)^2 = 59/4 T2 = 3.84 sec.The motorcycle will catch up with the car 3.84 seconds after it has accelerated.To determine the distance the motorcycle travelled from the time it accelerated,M = 46(T2) + 4(T2)^2M = 46(3.84) + 4(3.84)^2M = 176.64 + 58.98M = 235.62 m...Show more