5. a car with a velocity of 22m/s is accelerated uniformly at the rate of 1.6m/s² for 6.8s. what is its final velocity?

answer:

time needed: 8.76 s

speed of car: 18.4 m/s

explanation:

full question

at the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.1 m/s2.at the same instant a truck, traveling with a constant speed of 9.2 m/s, overtakes and passes the automobile.

(a) how far beyond the traffic signal will the automobile overtake the truck?

(b) how fast will the automobile be traveling at that instant?

this one is pretty straightforward. you know that at t=0 and x=0, the car starts moving with a constant acceleration of 2.1 m/s2.

at the same time, a truck moving at a speed of 9.2 m/s passes the car.

the idea here is that the distance they travel until the car overtakes the truck is the same for both the car, and the truck.

the same can be said for the time that passes until the two meet again.

so, if the car takes a time t to reach the truck, and both travel a distance d before that happens, then you an say that

    d=v⋅t→ for the truck

and

    d=v0=0⋅t+12⋅a⋅t2→ for the car

this will get you

    v⋅t

=12⋅a⋅t2

⇒t=2⋅va

t=2⋅9.2m

s2.1ms2

    =8.76 s

this means that the distance is

    d=v⋅t=9.2ms

⋅8.76s

    =80.6 m

the speed of the car is

    vcar=a⋅t=2.1ms2

⋅8.76s=18.4 m/s

explanation:

hope this you : ))

the maxium height is 34m/s