A car is initially at rest on a straight road. The graph shows the acceleration

Question

A car is initially at rest on a straight road. The graph shows the acceleration of the car along that road as a function of time. 0 1 2 3 4 5 6 7 89 10 11 12 13 14 t (s) What is the speed of the car at t-7 s? Submit Answer Incorrect. Tries 3/12 Previous Tries What distance does the car cover in the first 8 seconds? Submit Answer Tries 0/12 What is the speed of the car at t-14 s? Submit Answer Tries 0/12 Determine the distance covered by the car between t-11 s and t-14 s? Submit Answer Tries 0/12

Explanation / Answer

1. The velocity of the car after 1s is 0, because acceleration for 1s is 0

for t=1s to t=3s, a=2m/s2, t'=3-1=2s

so v=0+at'=2*2=4m/s

for t=3 to 4s t''=4.3=1s, a'=3m/s^2

then v'=v+a't''=4+3*1=7m/s

for t=4 to 5s acceleration is 0. So no change in velocity

for t=5s to 7s, a''=(-2)m/s^2, t'''=7-5=2s

vs=v'+at'''=7-2*2=7-4

So, at t=7s speed is vs=3m/s

another method is by area, for t=1s to 4s area is A1=2*2+3*1=7m/s

and for t=5s to 7s area is A2= -2*2= -4m/s

So, at t=7s v=A1+A2=7-4=3m/s

2. distance traveled in 8s

distance traveled in first second is 0

for time period of 1s to 3s

means t=2s, a=2m/s^2, u=0m/s traveled distance is

d1=ut+at2/2=0+2*22/2, d1=4m

for time period of 3s to 4s

at time 3s v=4m/s, a=3m/s2, t=4-3=1s

d2=vt+at2/2=4*1+3*12/2, d2=5.5m

for time period of 4s to 5s

velocity is constant v=7m/s, t=1s

So, d3=vt=7*1=7m

for time period of 5s to 7s

at time 5s, v=7m/s, a=(-2)m/s, t=7-5=2s

d4=vt+at2/2=7*2-2*22/2, d4=10m

for time period of 7s to 8s

at 7s, v=3m/s, a=3m/s2, t=8-7=1s

d5=3*1+3*12/2, d5=4.5m

so total distance traveled by car in 8s

d=d1+d2+d3+d4+d5=4+5.5+7+10+4.5=31m

3. For time period 7s to 8s

at 7s v=3m/s, a=3m/s2, t=8-7=1s

velocity after 8s is v1=v+at=3+3*1=6m/s

for time period 8s to 9s

  at 8s v1=6m/s, a=2m/s2, t=9-8=1s

  velocity after 9s is v2=v1+at=6+2*1=8m/s

for time period 9s to 11s

  at 9s v1=(-1)m/s, a=2m/s2, t=11-9=2s

  velocity after 9s is vf=v2+at=8-1*2=6m/s

after this acceleration is 0 so,

at t=14s speed is vf=6m/s

Another method is by area, at t=7s v=3m/s

for t=7s to 9s area is A3=3*1+2*1=5m/s

for t=9s to 11s area is A4= -1*2= -2m/s

So, velocity at t=14s is vf=v+A3+A4=3+5-2=6m/s

4. At t=11s v=6m/s and acceleration is 0 for t=11s to 14s

so, the distance traveled in this time period is d=v(14-11)=6*3=18m

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