A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by the equation x(t)=αt2−βt3, where α=1.50 ms2 and β=0.0500 ms3 Calculate the average velocity of the car for each time interval: a) t=0 to t=2.00s b) t=0 to t=4.00 s c) t=2.00s to t=4.00s

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A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by the equation x(t)=αt2−βt3, where α=1.50 ms2 and β=0.0500 ms3  Calculate the average velocity of the car for each time interval:  a) t=0 to t=2.00s  b) t=0 to t=4.00 s  c) t=2.00s to t=4.00s

chumants6g · Accepted Answer

Step 1  First we have to calculate the distance x at each t.  x(0)=Zero m  x(2)=1.50(2)2−0.0500(2)3=5.6m  x(4)=1.50(4)2−0.0500(4)3=20.8m  The average velocity is the displacement divided by the time interval at which this displacement happened.  vx, avg=ΔxΔt  Δx=xf−xi  Δt=tf−ti  Step 2  a) vx, avg=xf−xitf−ti  x(0)=Zero m  x(2)=1.50(2)2−0.0500(2)3=5.6m  vx, avg=5.6−02−0  =2.8ms  Step 3  b) x(0)=Zero m  x(4)=1.50(4)2−0.0500(4)3=20.8m  vx, avg=20.8−04−0  =5.2ms  Step 4  c) x(2)=1.50(2)2−0.0500(2)3=5.6m  x(4)=1.50(4)2−0.0500(4)3=20.8m  vx, avg=20.8−5.64−2  =7.6ms

Mary Herrera · Accepted Answer

Step 1  a) The equation for cars

nick1337 · Accepted Answer

Step 1 a) The position of the car as a function of time t is given by x(t)=αt2−βt3 where α=1.50m/s2β=0.05m/s3 The average velocity is given by the ratio between the displacement and the time taken: v=ΔxΔt The position at t = 0 is: x(0)=α×02−β×03=0 The position at t = 2.00 s is: x(2)=α×22−β×23=5.6m So the displacement is Δx=x(2)−x(0)=5.6−0=5.6m The time interval is Δt=2.0s−0s=2.0s And so, the average velocity in this interval is v=5.6m2.0s=2.8m/s b) The position at t = 0 is: x(0)=α×02−β×03=0 While the position at t = 4.00 s is: x(4)=α×42−β×43=20.8m So the displacement is Δx=x(4)−x(0)=20.8−0=20.8m The time interval is Δt=4.00−0=4.0s So the average velocity here is v=20.84.0=5.2m/s c) The position at t = 2 s is: x(2)=α×22−β×23=5.6m While the position at t = 4 s is:  x(4)=α×42−β×43=20.8m So the displacement is Δx=20.8−5.6=15.2m While the time interval is Δt=4.0−2.0=2.0s So the average velocity is v=15.22.0=7.6m/s

A Honda Civic travels in a straight line along a road. The car’s dista

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