Hello, Can I get a help with these problem? Please show the steps, formulas used

Question

Hello,

Can I get a help with these problem?

Please show the steps, formulas used and write clearly

An object moves with constant acceleration 4.05 m/s2 and over a time interval reaches a final velocity of 10.6 m/s.

(a) If its initial velocity is 5.3 m/s, what is its displacement during the time interval?
m

(b) What is the distance it travels during this interval?
m

(c) If its initial velocity is -5.3 m/s, what is its displacement during the time interval?
m

(d) What is the total distance it travels during the interval in part (c)?
m

Explanation / Answer

Dear student

Solution

Given quantities are

Acceleration a= 4.05 m/s2, final velocity vf=10.6 m/s2

a. Initial velocity vi=5.3 m/s2

To find time t

t=( vf-vi)/a=(10.6-5.3)/4.05=1.31 sec

Displacement S= vit+(1/2)at2

S= 5.3*1.31+(1/2)*4.05*(1.31)2

S=10.42 m

Distance is same thing as displacement here, because S & gt; 0 . In general distance is scalar quantity and displacement has direction, it is vector quantity
b. Hence distance travelled=10.42 m

c. Now initial velocity vi=-5.3 m/s2

Time t=(10.6-(-5.3))/4.05=(10.6+5.3)/4.05=3.93 sec

Displacement S= vit+(1/2)at2=-5.3*3.93+(1/2)*4.05*(3.93)2

Displacement S= 10.45 m

d. Total distance travelled in part will be initial displacement in -ve direction plus 10.45 m which is (20.83+10.45)=32.8 m

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.