Question

A car accelerates from rest at −3.00 m/s2. a. What is the velocity at the end of 5.0 s? b. What is the displacement after 5.0 s?

Answer 1

according to motion 1st equation
vf=vi+at
=0+-3×5
=-15m/s
.....
according to motion 2nd equation
d=vit+1/2at^2
=0+1/2×-3×5^2
=-37.5m

Answer 2

a. The velocity at the end of 5.0 seconds is -15 m/s.

b. The displacement after 5.0 seconds is -37.5 meters.

Given the following data:

• Acceleration = $-3 \;m/s^2$

• Initial velocity = 0 m/s (since the airplane starts from rest).

• Time = 5 seconds

a. To find the velocity at the end of 5.0 seconds, we would use the first equation of motion;

$V = U + at$

Where:

• a is the acceleration.

• V is the final velocity.

• U is the initial velocity.

• t is the time measured in seconds.

Substituting the given parameters into the formula, we have;

$V = 0 + (-3)(5)\\\\V = 0 - 15$

Final velocity, V = -15 m/s

Therefore, the velocity at the end of 5.0 seconds is -15 m/s.

b. To find the displacement after 5.0 seconds, we would use the second equation of motion:

$S = ut + \frac{1}{2} at^2$

Where:

• S is the distance travelled or displacement.

• u is the initial velocity.

• a is the acceleration.

• t is the time measured in seconds.

Substituting the parameters into the formula, we have;

$S = 0(t) + \frac{1}{2} (-3)(5^2)\\\\S = 0 - 1.5(25)\\\\S = 0 -37.5$

Displacement = -37.5 meters.

Therefore, the displacement after 5.0 seconds is -37.5 meters.

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