a truck moving a speed of 54km/h decelerates uniformly on applying brakes and its speed becomes 36km/h in 5sec. what is the distance covered by it in 5sec.
Answers
given :-
initial velocity = u = 54km/h
in m/s = 54 × 5/18 = 3 × 5 = 15m/s
final velocity = v = 36km/h
in m/s = 36 × 5/18 = 2 × 5 = 10m/s
time = t = 5 secs
acceleration = a = ?
so let's find the acceleration (deceleration) first
by equation of motion 1 we get,
➡ v = u + at
➡ 10 = 15 + 5a
➡ 10 - 15 = 5a
➡ -5 = 5a
➡ a = -5/5
➡ a = -1m/s²
hence, it's deceleration is -1m/s²
now, using second equation of motion
➡ s = ut + 1/2 at²
➡ s = (15 × 5) + 1/2 (-1) (5)²
➡ s = 75 + 1/2 (-25)
➡ s = 75 + (-25/2)
➡ s = 75 - 12.5
➡ s = 62.5m
the distance covered by it in 5 secs is 62.5m
Answer:
62.5m
Explanation:
we have the equation v= u+ at
v=36km/h=36*5/18=10m/s
u=54km/h=54*5/18=15m/s
t=5sec
applying it
10=15+5a
ie a=-1m/s^2
then we have the equation
v^2-u^2=2as
ie. 100-225=-2s
s=-125/-2=62.5m