Part a consider another special case in which the inclined plane is vertical (θ=π/2). in this case, for what value of m1 would t
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Answer
given,
v = (6 t - 3 t²) m/s
we know,
position of the particle
integrating both side
x = 3 t² - t³
Position of the particle at t= 3 s
x = 3 x 3² - 3³
x = 0 m
now, particle’s deceleration
a = 6 - 6 t
at t= 3 s
a = 6 - 6 x 3
a = -12 m/s²
distance traveled by the particle
x = 3 t² - t³
at t = 0 x = 0
t = 1 s , x = 3 (1)² - 1³ = 2 m
t = 2 s , x = 3(2)² - 2³ = 4 m
t = 3 s , x = 0 m
total distance traveled by the particle
D = distance in 0-1 s + distance in 1 -2 s + distance in 2 -3 s
D = 2 + 4 + 2 = 8 m
average speed of the particle
The braking distance is given by
Explanation:
When the driver of a car hits the pedal of the brakes, the car starts decelerating until it stops. Assuming the deceleration is constant, then the motion is a uniformly accelerated motion, so we can use the following suvat equation:
where
u is the initial speed of the car
v is the final speed of the car, which is zero because the car comes to rest:
v = 0
a is the acceleration of the car
s is the distance travelled by the car during the deceleration, so it is the braking distance
Therefore, re-arranging the equation for s, we find an expression for the braking distance:
Note that the sign of is negative since the car is decelerating, therefore the final sign of
is positive.
Learn more about accelerated motion:
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