we assume the acceleration is constant. we choose the initial and final points 1.40s apart, bracketing the slowing-down process. then we have a straightforward problem about a particle under constant acceleration. the initial velocity is v xi =632mi/h=632mi/h( 1mi 1609m )( 3600s 1h )=282m/s (a) taking v xf =v xi +a x t with v xf =0 a x = t v xf −v xf = 1.40s 0−282m/s =−202m/s 2 this has a magnitude of approximately 20g (b) similarly x f −x i = 2 1 (v xi +v xf )t= 2 1 (282m/s+0)(1.40s)=198m
Answer:
180^2 + 390^2 = force ^2 (Pythagoras) root of force^2 = 429.5N approx resultant force Acceleration = Force/Mass 429.5/270 = 1.5907 ms^-2 in a Southwesterly direction.
Explanation:
Initial velocity of the object: 5 m/s
Explanation:
The figure in the problem is missing: find it in attachment.
The graph in the figure represents the velocity of an object (v) versus the time passed (t).
Here we are asked to find the initial velocity of the object.
This means that we have to find the velocity of the object when the time is zero, so when
t = 0
By looking at the corresponding value on the y-axis (velocity), we see that when t = 0, then
v = 5 m/s
Therefore, the initial velocity of the object is 5 m/s.
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The force is known as electrostatic force (or electric force)