A sport car traveling eastward at 27.8 m/s slows at a constant rate to a stop in 8 minutes, what is the displacement of the sports car in this time interval?
Answers
Given info : A sport car traveling eastward at 27.8 m/s slows at a constant rate to a stop in 8 minutes.
To find : the displacement of the sports car in this time interval.
Solution : initial speed of car, u = 27.8 m/s
Final velocity of car, v = 0 [ because car stops finally]
Time taken to stop the car, t = 8 min = 8 × 60 = 480 sec
using formula,
v = u + at
⇒0 = 27.8 + a × 480
⇒a = -27.8/480 ≈ -0.06 m/s²
Now displacement of the sport car is given by, v² = u² + 2as
Here, v = 0, u = 27.8 , a = -0.06 m/s²
so, s = (27.8)²/2(0.06) = 6440.33 m
Therefore the displacement covered by the sport car is 6440.33m
Explanation:
Given info:-
A sport car travelling eastward at 27.8 m/s slows at a constant rate to stop in 8 mins.
Therefore, u=27.8 m/s.
t=8 mins =8×60=480 seconds.
v=0 m/s.
a=v-u/t=0-27.8/480=-27.8/480
To find:
Displacement of car in this time interval.
Formula used:
S=ut +1/2at^2.
Calculation:
=S=ut+1/2at^2
=S=27.8×480+1/2×27.8/480×480^2
=S=13344+0.03×23040
=S=13344.03×23040=307446451.2 m
HENCE DISPLACEMENT =307446451.2 m