Respuesta :
Answer:
a) t = 3.81s
b) distance travelled = 145.7m
A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle are both traveling at the same speed of 23.0 m/s , and the distance between them is 58.0 m . After t1 = 5.00 s , the motorcycle starts to accelerate at a rate of 8.00 m/s2 .
a. The motorcycle catches up with the car at some time t2. How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car? In other words, find t2−t1.
b. How far does the motorcycle travel from the moment it starts to accelerate (at time t1) until it catches up with the car (at time t2)? Should you need to use an answer from a previous part, make sure you use the unrounded value.
Explanation:
Given:
Initial speed of motorcycle and car = 23.0m/s
Acceleration of motorcycle = 8.00m/s^2
Distance between them = 58m
The distance covered by the motorcycle as the acceleration start can be given as;
d1 = ut + 0.5at^2. ......a
The distance from the point at which the motorcycle starts acceleration to the current position of the car at time t is;
d2 = 58 + ut
When they meet d1 equals d2
d1 = d2
ut + 0.5at^2 = 58 + ut
Substracting ut from both sides, we have;
0.5at^2 = 58 ......1
From equation 1.
t =√(58/0.5a)
a = 8 m/s2
t = √(58/4)
t = 3.81s
Therefore, d1 can be calculated using equation a
d1 = ut + 0.5at^2
d1 = 23(3.81) + 0.5(8)×3.81^2
d1 = 145.7m
