Find the distance travelled by a car moving with acceleration given by a(t)=t^2 + t, if it moves from t = 0 sec to t = 10 sec, if velocity of a car at t = 0 sec is 40 km/hr. *
Answers
Given : acceleration given by a(t)=t^2 + t, velocity of a car at t = 0 sec is 40 km/hr. *
To find : distance travelled by a car from t = 0 sec to t = 10 sec
Solution:
acceleration given by a(t)=t² + t,
a = dv/dt
=> dv = adt
=> dv = (t² + t)dt
integrating both sides
=> v = t³/3 + t²/2 + c
velocity of a car at t = 0 sec is 40 km/hr = 40 * 5/18 = 100/9 m/s
=> 100/9 = c
=> v = t³/3 + t²/2 + 100/9
Distance
d = t⁴/12 + t³/6 + 100t/9 + c
t = 0 sec to t = 10 sec
distance = 10⁴/12 + 10³/6 + 100*10/9 - 0
= 833.33 + 166.66 + 111.11
= 1,111.11 m
distance travelled by car = 1,111.11 m or 1.11 km
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Answer:
783.3km
Step-by-step explanation:
We know that,
a= ∫ ∫ (t^2+t)dtdt=∫[(t^3/3+t^2/2)+c]dt
= ∫100[(t^3/3+t^2/2)+40]dt
= 10003+50+400= 783.3km
hope it helps!.