Aspeedboat starts from rest and accelerates at - 2.01 m/s2 for 7.00 s. at the end of this time, the boat continues for an additional 6.00 s with an acceleration of +0.518 following this, the boat accelerates at 1.49 m/s2 for 8.00 s (a) what is the velocity of the boat at t = 21.0 s? (b) find the total displacement of the boat.

(a) 52.724 m/s

(b) Total displacement, d = 551.25 m

Solution:

As per the question:

Initial acceleration of the speed boat, a = - 2.01

Time duration, t = 7.00 s

Additional time, t' =6.00 s

Acceleration for additional time, a' = 0.518

The followed up acceleration, a'' = 1.49

Time duration, t'' = 8.00 s

(a) Now, to calculate the velocity of the boat at timer, t = 21.0 s, we have:

After the initial 7.00 s, the velocity of the boat, from eqn-1 of motion:

v = u + at

v = 0 - 2.1(7.00) = - 14.7 m/s

After t + t' = 13 s:

v' = v + at

v' =  14.7 + 0.518(13) =  21.434 m/s

Now, velocity of the boat after t = 21 s:

v'' = v' + a''t

v'' = 21.434 + 1.49(21) = 52.724 m/s

(b) Now, the total displacement, d:

For the first case:

d = ut +

For the second case:

d = v't' = 21.434(6) = 128.6 m

For the third case:

d = ut +

For the fourth case:

d = v''t'' = 52.724(8) = 421.79 m

For the last case:

d = ut +

Total displacement, d = -51.45 + 128.6 + 4.65 + 421.79 + 47.68 = 551.25 m

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.

electric field is a function of 1/r^2

yes, the force per unit charge for a point charge is inverse square law.