Answer:
37.125 m
Explanation:
Using the equation of motion
s=ut+0.5at^{2} where s is distance, u is initial velocity, t is time and a is acceleration
<u>Distance during acceleration</u>
Acceleration, a=\frac {V_{final}-V_{initial}}{t} where V_{final} is final velocity and V_{initial} is initial velocity.
Substituting 0.0 m/s for initial velocity and 4.5 m/s for final velocity, acceleration will be
a=\frac {4.5 m/s-0 m/s}{4.5 s}=1 m/s^{2}
Then substituting u for 0 m/s, t for 4.5 s and a for 1 m/s^{2} into the equation of motion
s=0*4.5+ 0.5*1*4.5^{2}=0+10.125
=10.125 m
<u>Distance at a constant speed</u>
At a constant speed, there's no acceleration and since speed=distance/time then distance is speed*time
Distance=4.5 m/s*6 s=27 m
<u>Total distance</u>
Total=27+10.125=37.125 m
Answer:
s = 589.3 m
Explanation:
Let the truck and car meet at a distance = s m
The truck is moving at constant velocity = v
so s= v * t ---------- (1)
car:
Vi = 0 m/s
a = 3.9 m/s²
s = Vi* t + 1/2 a t²
s= 0 * t + 1/2 a t²
s = 1/2 a t² ----------- (2)
compare equation (1) and equation (2)
s= v * t = 1/2 a t²
⇒ v * t = 1/2 a t²
⇒ t = 2 * v/ a
⇒ t = (2 * 33.9 )/ 3.9
⇒ t = 17. 38 s
Now
from equation (1)
s= v * t
s= 33.9 * 17.38
⇒ s = 589.3 m
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Acceleration=velocity/time
acceleration=28/4.22
therefore, acceleration=6.64
