At time t=0, a scooter has a velocity of 21 meter per sec. It slows down with an acceleration of 50 cm per sec square. At the end of 4 sec , the scooter has travelled :
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At the end of 4 sec the scooter would have traveled a distance of 80 meters.
Explanation:
from the laws of motion equations we know that,
v = u + at
where v = final velocity
u = initial velocity = 21m/s
a = -50 cm/s² = - 0.5 m/s²
t = time = 4 sec
Hence
v = 21 - 0.5 x 4 = 19 m/s
Also ,
v² - u² = 2as
=> 19² - 21² = 2 x -0.5 x s
=> s = 21² - 19² = 441 - 361 = 80 meters
Hence at the end of 4 sec the scooter would have traveled a distance of 80 meters.
Answered by
11
at time t = 0, the scooter has a velocity of 21m/s
e.g., initial velocity of scooter , u = 21 m/s
scooter slows down with an acceleration of 50cm/s² .
e.g., acceleration of scooter , a = -50cm/s² = -0.5m/s² [ here we considered negative sign because question says scooter slows down ]
now, use formula , s = ut + 1/2 at² to find distance travelled in 4 sec by scooter .
s = 21 × 4 + 1/2 × -0.5 × 4²
= 84 - 1/2 × 0.5 × 16
= 84 - 4
= 80m
hence, scooter travelled 80m in 4 sec.
e.g., initial velocity of scooter , u = 21 m/s
scooter slows down with an acceleration of 50cm/s² .
e.g., acceleration of scooter , a = -50cm/s² = -0.5m/s² [ here we considered negative sign because question says scooter slows down ]
now, use formula , s = ut + 1/2 at² to find distance travelled in 4 sec by scooter .
s = 21 × 4 + 1/2 × -0.5 × 4²
= 84 - 1/2 × 0.5 × 16
= 84 - 4
= 80m
hence, scooter travelled 80m in 4 sec.
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