Amotorcycle 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 22.0 m/s , and the distance between them is 60.0 m . after t1= 5.00 s , the motorcycle starts to accelerate at a rate of 5.00 m/s^2 . the motorcycle catches up with the car at some time . 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. 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.

a) 4.9 s

b) 167.8 m

Explanation:

Hello!

To solve this question we need to make use of the equations of motion of both the motorcycle xm(t) and the car xc(t) at t=5

Let us consider the position of the motorcycle at t=5 as the origin, that is:

xm(t+5) = vt + (1/2)at^2

xc(t+5) = vt + 60 m

where v = 22.0m/s  and   a=5m/s^2

We are looking for the time t' when the position of the car and the motorcycle are the same:

xm(t'+5)=xc(t'+5)

vt' + (1/2)at'^2 = vt' +60m

t' = √(120 m /a) = 4.89898... s

Since we are considering the origin of the cooordinate system at the position when the motorcycle starts to accelerate, the distance travelled by the motorcycle until it catches the car is given by:

xm(t'+5)= vt' + (1/2)at'^2

xm(9.89898s) = (22 * 9.89898 + 2.5 * 9.89898^2)m

xm(9.89898s)= 167.777... m

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle