1\ Acar is moving with acceleration 200 m in 50 sec reached to velocity 10 m\sec .what the total distance the car moved in 60 sec
There is no reason to keep posting the same question, unless you want to be ignored.
orry i want to be ignored
Amotorcycle slowed down to rest from avelocity of 10m\sec whiletraveling 30m what is its average acceleration
Esteemed professor I hope to resolve the issue as soon as possible and thank you
Amotorcycle slowed down to rest from avelocity of 10m\sec whiletraveling 30m what is its average acceleration
Motorcycle slowed down to rest from avelocity of 10 m\sec while traveling 30 m what is its average acceleration
\ A Motorcycle slowed down to rest from avelocity of 10m\sec while traveling 30 m. what is its average acceleration
To find the total distance the car moved in 60 seconds, we can break it down into two parts:
1. Calculate the distance covered during the first 50 seconds when the car reaches a velocity of 10 m/s.
2. Calculate the distance covered during the last 10 seconds when the car maintains a constant velocity of 10 m/s.
Let's calculate each part step by step:
1. Calculate the distance covered during the first 50 seconds:
The car starts from rest, with an acceleration of 200 m/s^2, and reaches a velocity of 10 m/s in 50 seconds. We can use the formula:
v = u + at,
where:
v = final velocity (10 m/s),
u = initial velocity (0 m/s),
a = acceleration (200 m/s^2),
t = time taken (50 s).
Rearranging the formula to solve for the initial velocity (u), we get:
u = v - at.
Plugging in the values, we have:
u = 10 m/s - (200 m/s^2 * 50 s)
= 10 m/s - 10000 m/s^2
= -9990 m/s.
The negative sign indicates that the initial velocity was in the opposite direction of the final velocity. However, since we are considering distance, not displacement, we can ignore the negative sign.
Now, we can use the formula:
s = ut + 0.5at^2,
where:
s = distance,
u = initial velocity,
t = time taken,
a = acceleration.
Plugging in the values, we get:
s = (-9990 m/s) * (50 s) + 0.5 * (200 m/s^2) * (50 s)^2
= -499500 m + 0.5 * 200 m/s^2 * 2500 s^2
= -499500 m + 250000 m
= -249500 m.
Again, we ignore the negative sign when considering distance, so the distance covered during the first 50 seconds is:
distance_1 = 249500 m.
2. Calculate the distance covered during the last 10 seconds:
During this period, the car maintains a constant velocity of 10 m/s, so we can use the equation:
distance_2 = velocity * time
= 10 m/s * 10 s
= 100 m.
Finally, we add the distances covered during the two periods to get the total distance covered in 60 seconds:
total distance = distance_1 + distance_2
= 249500 m + 100 m
= 249600 m.
Therefore, the car moved a total distance of 249600 meters in 60 seconds.