Acar traveling initially at +7.9 m/s accelerates uniformly at the rate of +0.72 m/s2 for a distance of 265 m. what is its velocity at the end of the acceleration? answer in units of m/s. 10.0 points what is its velocity after it accelerates for 114 m? answer in units of m/s

in first case velocity =21.07 m/sec in second case velocity =15.05 m/sec

Explanation:

We have given initial velocity u = 7.9 m/sec

Acceleration

Distance S = 265 m

Now according to third law of motion here v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance

So

v = 21.07 m/sec

In second case s =114 m

So

v =15.05 m/sec

1. 21.07 m/s

2. 15.05 m/s

Given:

initial speed of the car, u = 7.9 m/s

distance covered by the car, d = 265 m

acceleration, a = 0.72

Solution:

To calculate the velocity at the end of acceleration, we use the third eqn of motion:

Now,

Velocity after it accelerates for a distance for 114 m:

Here d = 114 m

Again, from third eqn of motion:

the amount of current in the coil

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2