A cart started at $t = 0$ , its acceleration varies with time as shown in figure. Find the distance travelled in $30$ seconds and draw the position-time graph.

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Solution

$u = 0, a = 10 m / s^{2}$
$s = u t + \frac{1}{2} a t^{2}$
$s_{1} = 0 + \frac{1}{2} \times 10 \times 10^{2}$
$s_{1} = 500 m$
Now,from t=0 to 20s,
$a = 0, v = u + a t$
$v = 0 + 10 \times 10 = 100 m / s$
$s_{2} = v t = 100 \times (20 - 10)$
Now, $a = - 10 m / s$
$s_{3} = u t - \frac{1}{2} a t^{2}$
$= (100) \times 10 - \frac{1}{2} \times 10 \times 100$
$s_{3} = 500 m$
Total distance = $500 + 1000 + 500 = 2000 m$

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A cart started at t=0, its acceleration varies with time as shown in figure. Find the distance travelled in 30 seconds and draw the position-time graph.

u=0,a=10m/s2s=ut+12at2s1=0+12×10×102s1=500mNow,from t=0 to 20s,a=0,v=u+atv=0+10×10=100m/ss2=vt=100×(20−10)Now, a=−10m/ss3=ut−12at2=(100)×10−12×10×100s3=500mTotal distance =500+1000+500=2000m

A cart started at $t = 0$ , its acceleration varies with time as shown in figure. Find the distance travelled in $30$ seconds and draw the position-time graph.