Consider a particle initially moving with a velocity of 5m/s starts deaccelarating at a constant rate of 2m/s*2 find the distance travelled in 2sec? Ans:2m
d=vi*t-1/2 a t^2=5*2-1*4=6m
To find the distance traveled by the particle in 2 seconds, we need to use the equation of motion for constant acceleration. The equation is:
d = ut + (1/2)at^2
Where:
d = distance traveled
u = initial velocity
a = acceleration
t = time
Given:
u = 5 m/s (initial velocity)
a = -2 m/s^2 (acceleration, negative sign indicates deceleration)
t = 2 seconds (time)
Plugging the values into the equation, we get:
d = (5 * 2) + (1/2)(-2)*(2^2)
= 10 + (-2)
= 8 meters
So the particle would have traveled a distance of 8 meters in 2 seconds. Therefore, the given answer of 2 meters is incorrect.