Question

please solve the following​

Answer 1

Answer:

1) (i) 15 m/s²

1) (ii) t = 10 to t = 14

1) (iii) 20 m/s²

1) (iv) 90 metres

2) (a) 12 m/s

2) (b) 720 metres

Explanation:

1) (i) Average acceleration is given by $\Delta v\over\Delta t$

Av. acc. = ${30\over2}=15\,\,m/s^2$

1) (ii) When acceleration becomes zero, then velocity becomes constant, i.e., the v-t graph runs parallel to time axis. So, between time 10 to 14 seconds, acceleration is zero.

1) (iii) Maximum acceleration corresponds to the steepest slope. In graph, we can see that steepest slope is between t = 14 sec to t = 16 sec. In this time interval of 2 second, velocity comes from 40 to 0 m/s.

Here, acceleration is, ${\Delta v\over\Delta t}={-40\over2}=-20\,\,m/s^2$

Please note that, here, the magnitude of acceleration is maximum in the entire graph, but it's in negative sign. If you want to have value maximum integer value of acceleration, then you will find it between time interval t = 6 sec to t = 8 sec. Here, acceleration is ${\Delta v\over\Delta t}={{60-30}\over2}={30\over2}=15\,\,m/s^2$

1) (iv) Displacement is given by the area of the v-t graph. So displacement from t = 0 to t = 6 seconds will be, ${1\over2}\times30\times6=90\,\,metres.$

2) acceleration, a = 0.1 m/s²

   time, t = 2 minutes = 120 seconds.

(a) ∵ $v=u+at$

∴ $v=0+(0.1\times120)=12\,\,m/s$

(b) ∵$s=ut+{1\over2}at^2$

∴$s=0+{1\over2}\times0.1\times(120)^2=720\,\,metres$