Question #717

If a car starts from rest and accelerates at 5.0 m/sec^2 for 6.0 seconds, how far did it travel and how fast was it moving at the end of the 6.0 seconds?
Can you show work please?

Expert's answer

The motion of the car in that 6 seconds have been a motion with constant acceleration. One of the

equations of motion of a body in constant acceleration is:

a=Δv/Δt⇒Δv=a*Δt

⇒v-v0=a*Δt⇒v=a*Δt+v0

Where v is the final velocity and v0 is the initial velocity and v0 = 0. Now with all the data we have, we can get for the final velocity:

v=a*Δt=5*6=30m/s

For the distance the car traveled in that time, we use another one of the equations of motion of a body in constant acceleration, namely the time-independent one:

v²-v²0=2a*Δx⇒Δx=(v²-v²0)/2a

and since v0 = 0 we get Δx=v²/2a

and using the data we have, we get Δx=v²/2a=30²/2*5=900/10=90m

We could nd the distance by another method, using another equation of motion:

Delta;x=x-x0=1/2a*t²+v0*t=1/2*5*6²+0=90m

equations of motion of a body in constant acceleration is:

a=Δv/Δt⇒Δv=a*Δt

⇒v-v0=a*Δt⇒v=a*Δt+v0

Where v is the final velocity and v0 is the initial velocity and v0 = 0. Now with all the data we have, we can get for the final velocity:

v=a*Δt=5*6=30m/s

For the distance the car traveled in that time, we use another one of the equations of motion of a body in constant acceleration, namely the time-independent one:

v²-v²0=2a*Δx⇒Δx=(v²-v²0)/2a

and since v0 = 0 we get Δx=v²/2a

and using the data we have, we get Δx=v²/2a=30²/2*5=900/10=90m

We could nd the distance by another method, using another equation of motion:

Delta;x=x-x0=1/2a*t²+v0*t=1/2*5*6²+0=90m

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